Physics · Secondary School · Gravitation

Gravitation

The apple that falls from the branch, the Moon that goes round the Earth once a month and the sea that rises and falls on the beach all share the same origin, the gravitational attraction between masses. We start from Newton's law, make sense of the gravitational field and of orbits, take apart the idea that there is no gravity in space, weigh planets with Kepler's laws and finish with escape velocity and the tides.

1Law of gravitation
2Gravitational field
3Orbits
4Weightlessness
5Kepler and masses
6Escape velocity
7Tides
✓Challenges

STEP 1

Does the force that brings down the apple hold up the Moon?

In the Circular Motion lesson, in the step on centripetal force, we saw that the Moon only curves round the Earth because some force pulls it towards the centre, and we said in advance that this force is gravity. We now take a closer look at the law Newton proposed for it, the law of universal gravitation.

Any two bodies attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Because the constant \(G\) is tiny, the attraction between two everyday objects tends to go unnoticed, and it only becomes large when at least one of the masses is the size of a planet.

Newton tested the idea with a famous calculation. The Moon is about 60 Earth radii from the centre of the Earth and, if the law holds, gravity there must be \(60^2 = 3600\) times weaker than here. In 1 s the apple falls 4.9 m, and the Moon should fall towards the Earth 3600 times less, a little over 1 mm, which is precisely the deviation the observed orbit requires.

The force comes in pairs, as Newton's 3rd law demands. The apple pulls the Earth just as hard as the Earth pulls the apple, and the only reason we do not see the Earth rise to meet it is that, with such a large mass, the Earth's acceleration is practically zero.

\(F = \dfrac{G\,M\,m}{d^2}\)\(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\)\(d\) is the distance between the centres of the bodies. Throughout the lesson we use \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\), the mass of the Earth \(M_E = 6 \cdot 10^{24}\ \text{kg}\) and the radius of the Earth \(R_E = 6400\ \text{km}\). In the simulation, sizes and distances are not to scale, and \(d_0\) is the starting distance of each scenario.

gravitational force on each body

Let's discuss

With the spheres, double the distance and then triple it. Our first intuition may be that the force drops to a half and to a third; check on the graph whether it drops by more than that.
Double mass A and then mass B. What happens to the force, and to the arrows on both sides?
Choose Earth and apple and compare the two accelerations. Why does only the apple seem to move?
In Earth and Moon, leave the distance at 60 radii and compare the Moon's acceleration with the apple's 9.8 m/s². Is the ratio close to 3600?

Mass A

Mass B

Distance

Force

Distance

Force compared with that at d₀

Acceleration of A

Acceleration of B

Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Two people, of 60 kg and 80 kg, are sitting 2 m apart. With what force do they attract each other? Treat each person as a point and use \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\).

Show solution
By the law of gravitation, \(F = \dfrac{G\,m_1\,m_2}{d^2}\) \(= \dfrac{6.7 \cdot 10^{-11} \cdot 60 \cdot 80}{2^2}\).
\(F \approx 8.0 \cdot 10^{-8}\ \text{N}\), a force far too small for us to feel.
basic
Two bodies attract each other with a force \(F\). If the distance between them triples, what does the force become?

Show solution
The force is inversely proportional to the square of the distance, and tripling \(d\) divides the force by \(3^2\).
The force becomes \(F/9\).
basic
A 0.2 kg apple falls from the tree, pulled by the Earth with a force of 2 N. With what force does the apple pull the Earth? Why do we not see the Earth rise to meet it?

Show solution
By Newton's 3rd law, the apple pulls the Earth with a force of the same magnitude, in the opposite direction.
The Earth's acceleration, however, is \(a = \dfrac{F}{M_E} = \dfrac{2}{6 \cdot 10^{24}}\) \(\approx 3 \cdot 10^{-25}\ \text{m/s}^2\), far too small for any measurement.
The apple pulls the Earth upwards with 2 N, and the Earth's enormous mass leaves its acceleration practically zero.
basic
Calculate the force with which the Earth attracts a 1 kg body at the surface. Use \(M_E = 6 \cdot 10^{24}\ \text{kg}\), \(R_E = 6400\ \text{km}\) and \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\).

Show solution
We convert the radius to metres, \(R_E = 6.4 \cdot 10^6\ \text{m}\), and use \(F = \dfrac{G\,M_E\,m}{R_E^2}\) \(= \dfrac{6.7 \cdot 10^{-11} \cdot 6 \cdot 10^{24} \cdot 1}{(6.4 \cdot 10^6)^2}\).
\(F \approx 9.8\ \text{N}\), the weight of 1 kg we already knew.
intermediate
The Moon is about 60 Earth radii from the centre of the Earth. How many times smaller is the acceleration caused by the Earth there than at the surface? How far does the Moon fall towards the Earth in 1 s, if at the surface a body falls 4.9 m in that time?

Show solution
The acceleration falls with the square of the distance, and at 60 radii it is \(60^2\) times smaller.
In 1 s, the fall is also 3600 times smaller, \(\dfrac{4.9\ \text{m}}{3600}\).
The acceleration is 3600 times smaller, and the Moon falls about 1.4 mm every second.
intermediate
Calculate the force of attraction between the Earth and the Moon. Use \(M_E = 6 \cdot 10^{24}\ \text{kg}\), \(M_M = 7.4 \cdot 10^{22}\ \text{kg}\), \(d = 3.84 \cdot 10^8\ \text{m}\) and \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\).

Show solution
In the law of gravitation, the numerator is \(G\,M_E\,M_M\) \(= 6.7 \cdot 10^{-11} \cdot 6 \cdot 10^{24}\) \(\cdot\ 7.4 \cdot 10^{22} \approx 2.97 \cdot 10^{37}\), and the denominator is \(d^2 = (3.84 \cdot 10^8)^2 \approx 1.47 \cdot 10^{17}\).
So \(F = \dfrac{2.97 \cdot 10^{37}}{1.47 \cdot 10^{17}}\).
\(F \approx 2 \cdot 10^{20}\ \text{N}\)
intermediate
Two bodies attract each other with a force \(F\). We double the mass of each one and also the distance between them. What is the new force?

Show solution
With the new quantities, \(F' = \dfrac{G\,(2m_1)(2m_2)}{(2d)^2}\) \(= \dfrac{4}{4} \cdot \dfrac{G\,m_1\,m_2}{d^2}\).
The force is still equal to \(F\).
intermediate
A spacecraft travels in a straight line from the Earth to the Moon. At what point do the pulls of the two cancel out? The mass of the Earth is about 81 times that of the Moon, and the distance between the centres is 384,000 km.

Show solution
Calling \(x\) the distance from the centre of the Earth, the forces are equal when \(\dfrac{G\,M_E\,m}{x^2} = \dfrac{G\,M_M\,m}{(d - x)^2}\).
This gives \(\dfrac{x}{d - x} = \sqrt{\dfrac{M_E}{M_M}} = \sqrt{81} = 9\), and so \(x = \dfrac{9}{10}\,d\).
About 346,000 km from the centre of the Earth, nine tenths of the way.
challenge
Newton checked the law of gravitation against the motion of the Moon, which follows an almost circular orbit of radius \(3.84 \cdot 10^8\ \text{m}\) in 27.3 days. Calculate the Moon's centripetal acceleration and compare it with \(g/60^2\), using \(g = 9.8\ \text{m/s}^2\) and remembering that this radius is about 60 Earth radii.

Show solution
The period in seconds is \(T = 27.3 \cdot 86\,400 \approx 2.36 \cdot 10^6\ \text{s}\).
The centripetal acceleration is \(a_c = \dfrac{4\pi^2\,r}{T^2} = \dfrac{4\pi^2 \cdot 3.84 \cdot 10^8}{(2.36 \cdot 10^6)^2}\) \(\approx 2.7 \cdot 10^{-3}\ \text{m/s}^2\).
The law of gravitation predicts \(\dfrac{9.8}{3600} \approx 2.7 \cdot 10^{-3}\ \text{m/s}^2\).
The two values agree, and the gravity that brings down the apple explains the Moon's orbit.
challenge
Two 100 kg lead spheres have their centres 0.5 m apart. Calculate the attraction between them and the mass of an object that, on Earth, would have a weight equal to this force. Use \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\) and \(g = 10\ \text{m/s}^2\).

Show solution
The attraction is \(F = \dfrac{6.7 \cdot 10^{-11} \cdot 100 \cdot 100}{0.5^2}\) \(\approx 2.7 \cdot 10^{-6}\ \text{N}\).
An equal weight corresponds to a mass \(m = \dfrac{F}{g} = \dfrac{2.7 \cdot 10^{-6}}{10}\).
\(m \approx 2.7 \cdot 10^{-7}\ \text{kg}\), less than a third of a milligram, which is why Cavendish needed a very sensitive torsion balance to measure this force.

STEP 2

How much does an astronaut weigh on another world?

We can think of the Earth as something that modifies the space around it. At each point there is a gravitational field \(\vec g\), the force the Earth would exert on each kilogram placed there, and the weight of a body is its mass times this field.

Dividing Newton's law by the mass \(m\), we arrive at \(g = G\,M/r^2\). With the lesson's numbers, the Earth's surface comes out with \(g \approx 9.8\ \text{m/s}^2\), the value already used in Kinematics and Dynamics, where it was nearly always rounded to 10.

Mass measures the amount of matter and the inertia of a body, and it does not change from one planet to another. Weight changes along with the local \(g\), and a set of bathroom scales, which measures force and divides by 9.8, would show a much smaller number on Mars than on Earth.

\(g = \dfrac{G\,M}{r^2}\)\(W = m\,g\)\(\dfrac{g}{g_0} = \left(\dfrac{R}{R + h}\right)^2\)\(r = R + h\) is the distance to the centre of the planet, \(R\) is its radius and \(g_0\) is the value at the surface. The values in the table are approximate and come from the masses and radii used in the simulation; Jupiter's applies at the cloud tops.

g at the surface
Body	Mass (kg)	Radius (km)	g (m/s²)
Mercury	3.3·10²³	2440	3.7
Venus	4.9·10²⁴	6050	8.9
Earth	6.0·10²⁴	6400	9.8
Moon	7.4·10²²	1740	1.6
Mars	6.4·10²³	3400	3.7
Jupiter	1.9·10²⁷	71,500	24.9

With height, \(g\) decreases with the square of the distance to the centre. At 400 km, where the International Space Station flies, the field is still \(g \approx 8.7\ \text{m/s}^2\), almost 90% of the value on the ground, something that may surprise anyone who pictured space without gravity.

Let's discuss

Weigh the astronaut on the Earth, on the Moon and on Jupiter. What changes in the scales reading, and what stays the same?
Launch the rocket on Earth and follow the graph. At what height does \(g\) drop to half, and at what height does it drop to a quarter?
Take the rocket to 400 km, the height of the ISS. What is \(g\) there, compared with the ground?
Compare the Moon's curve with the Earth's over the first thousand kilometres. In which of them does \(g\) seem to drop faster, and how can we explain this?

Person's mass

Altitude

g at the surface

g at this altitude

Mass

Weight

Scales reading

Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
On the Moon, \(g \approx 1.6\ \text{m/s}^2\). What is the weight of an 80 kg astronaut there, and what is their mass?

Show solution
The weight is \(W = m\,g = 80 \cdot 1.6\), and the mass does not change from one place to another.
\(W = 128\ \text{N}\), and the mass is still 80 kg.
basic
Calculate \(g\) at the surface of the Earth with \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\), \(M_E = 6 \cdot 10^{24}\ \text{kg}\) and \(R_E = 6400\ \text{km}\).

Show solution
\(g = \dfrac{G\,M_E}{R_E^2}\) \(= \dfrac{6.7 \cdot 10^{-11} \cdot 6 \cdot 10^{24}}{(6.4 \cdot 10^6)^2}\) \(= \dfrac{4.02 \cdot 10^{14}}{4.1 \cdot 10^{13}}\).
\(g \approx 9.8\ \text{m/s}^2\)
basic
A set of bathroom scales measures the force we exert on it and shows the result in kilograms, dividing by 9.8 m/s². A 70 kg person steps on these scales on Mars, where \(g \approx 3.7\ \text{m/s}^2\). What do the scales show? Has the person got any slimmer?

Show solution
On Mars, the force on the scales is the weight \(W = 70 \cdot 3.7 = 259\ \text{N}\), and the scales divide this value by 9.8, \(\dfrac{259}{9.8}\).
The scales show about 26 kg, and the mass is still 70 kg; what changed was the weight, along with \(g\).
basic
At an altitude equal to the Earth's radius, the distance to the centre doubles. What is \(g\) there, if at the surface it is 9.8 m/s²?

Show solution
With twice the distance to the centre, \(g\) drops to \(\dfrac{1}{2^2}\) of its surface value, \(\dfrac{9.8}{4}\).
\(g = 2.45\ \text{m/s}^2\)
intermediate
The International Space Station flies at a height of about 400 km. Calculate \(g\) at this height with the lesson's data, \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and \(R_E = 6400\ \text{km}\), and compare it with the value at the surface, 9.8 m/s².

Show solution
The distance to the centre is \(r = 6400 + 400 = 6800\ \text{km}\), and \(g = \dfrac{G\,M_E}{r^2} = \dfrac{4.02 \cdot 10^{14}}{(6.8 \cdot 10^6)^2}\).
Compared with the surface, the ratio is \(\left(\dfrac{6400}{6800}\right)^2 \approx 0.89\).
\(g \approx 8.7\ \text{m/s}^2\), about 89% of the value on the ground.
intermediate
A planet has twice the mass of the Earth and twice its radius. What is \(g\) at its surface, if on Earth it is 9.8 m/s²?

Show solution
\(g' = \dfrac{G\,(2M)}{(2R)^2} = \dfrac{2}{4} \cdot \dfrac{G\,M}{R^2}\).
\(g' = 4.9\ \text{m/s}^2\), half the Earth's value.
intermediate
At what altitude above the Earth's surface is \(g\) one ninth of its value on the ground? Give the answer in kilometres, with \(R_E = 6400\ \text{km}\).

Show solution
Since \(g \propto 1/r^2\), for \(g\) to fall to 1/9 the distance to the centre has to triple, \(r = 3\,R_E\).
The altitude is \(h = r - R_E = 2\,R_E\).
\(h = 12\,800\ \text{km}\)
intermediate
Mars has a mass of \(6.4 \cdot 10^{23}\ \text{kg}\) and a radius of 3400 km. Calculate \(g\) at the surface of Mars, with \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\).

Show solution
\(g = \dfrac{6.7 \cdot 10^{-11} \cdot 6.4 \cdot 10^{23}}{(3.4 \cdot 10^6)^2}\) \(= \dfrac{4.29 \cdot 10^{13}}{1.16 \cdot 10^{13}}\).
\(g \approx 3.7\ \text{m/s}^2\)
challenge
Henry Cavendish became known for weighing the Earth, by measuring \(G\) in the laboratory. Show how, knowing \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\), \(g = 9.8\ \text{m/s}^2\) and \(R_E = 6400\ \text{km}\), we calculate the mass of the Earth.

Show solution
From the expression for the field at the surface, \(g = \dfrac{G\,M_E}{R_E^2}\), we isolate \(M_E = \dfrac{g\,R_E^2}{G}\) \(= \dfrac{9.8 \cdot (6.4 \cdot 10^6)^2}{6.7 \cdot 10^{-11}}\).
\(M_E \approx 6.0 \cdot 10^{24}\ \text{kg}\)
challenge
A planet has the same density as the Earth and half its radius. What is \(g\) at its surface, compared with the Earth's? Remember that the volume of a sphere is \(\dfrac{4}{3}\pi R^3\).

Show solution
With density \(\rho\), the mass is \(M = \rho \cdot \dfrac{4}{3}\pi R^3\), and so \(g = \dfrac{G\,M}{R^2} = \dfrac{4}{3}\pi\,G\,\rho\,R\).
With the same density, \(g\) is proportional to the radius.
The planet's \(g\) is half the Earth's \(g\), about 4.9 m/s².

STEP 3

Why does the Moon not fall onto the Earth?

Back in the Circular Motion lesson, we launched a satellite and saw that, with the right speed, it goes round and round instead of falling. Gravity plays the part of the centripetal force there, and setting the two equal gives the speed of the circular orbit.

Newton pictured the same thing with a cannon on top of an extremely high mountain, above the air. A slow cannonball lands nearby, a faster one lands further away and, at a certain speed, the Earth's surface curves away from the ball as fast as the ball falls, so that it never reaches the ground. A satellite is falling all the time and, all the time, missing the Earth.

The satellite's mass cancels out in the calculation, and the speed and the period depend only on the mass of the Earth and on the radius of the orbit. The higher the satellite, the slower it moves and the longer it takes to go round, and the comparison between the ISS, which completes a lap in about 93 minutes, and the Moon, which takes 27 days, bears this out.

A special case is the geostationary satellite, which orbits above the equator, in the same direction as the Earth turns, with a period of 24 h. Seen from the ground, it stays still in the sky, which is why satellite TV dishes can be fixed in place. The calculation puts the radius of this orbit at about 42,000 km, some 36,000 km above the surface.

\(\dfrac{m\,v^2}{r} = \dfrac{G\,M\,m}{r^2}\)\(v = \sqrt{\dfrac{G\,M}{r}}\)\(T = 2\pi\sqrt{\dfrac{r^3}{G\,M}}\)\(r\) is the radius of the orbit, measured from the centre of the Earth, and \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\). In the simulation, the mountain is 400 km high and we assume there is no air; near the real surface, the air would slow the ball down long before. The path is calculated step by step, in steps of 5 s.

Let's discuss

Fire at 4 km/s and at 6 km/s and compare where the ball lands. Does the range seem to grow in the same proportion as the speed?
Look for the speed of the circular orbit and compare it with \(\sqrt{G\,M/r}\) for \(r = 6800\ \text{km}\).
Fire at 9 km/s and at 10 km/s. Where is the highest point of each ellipse, and which point does the ball always pass through again?
How long does the ball take to go once round the circular orbit? Compare it with the period of the ISS, about 93 minutes.

Speed

Firing speed

Circular speed there

Escape velocity there

Result

Period

Flight time

Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
What would the speed of a satellite be in a circular orbit skimming the Earth's surface, if there were no air and no mountains? Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and \(R_E = 6.4 \cdot 10^6\ \text{m}\).

Show solution
\(v = \sqrt{\dfrac{G\,M_E}{R_E}} = \sqrt{\dfrac{4.02 \cdot 10^{14}}{6.4 \cdot 10^6}}\) \(= \sqrt{6.28 \cdot 10^7}\).
\(v \approx 7.9\ \text{km/s}\)
basic
If gravity pulls the Moon towards the Earth all the time, why does the Moon not fall?

Show solution
The Moon does fall, in the sense that its path curves towards the Earth all the time.
At the same time, it moves sideways at about 1 km/s, and the fall only bends its path round the Earth, like the ball from Newton's cannon at orbital speed.
The Moon is always falling and always missing the Earth, and gravity plays the part of the centripetal force.
basic
Two satellites, one of 200 kg and the other of 2000 kg, travel in the same circular orbit. Which of them has the longer period?

Show solution
Setting the gravitational force equal to the centripetal force, \(\dfrac{m\,v^2}{r} = \dfrac{G\,M\,m}{r^2}\), the satellite's mass \(m\) cancels, and \(T = 2\pi\sqrt{r^3/(G\,M)}\) depends only on \(r\) and on the mass of the Earth.
Both have the same period.
basic
The ISS orbits 6800 km from the centre of the Earth. Calculate its speed and its period. Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\).

Show solution
\(v = \sqrt{\dfrac{4.02 \cdot 10^{14}}{6.8 \cdot 10^6}}\) \(\approx 7.69 \cdot 10^3\ \text{m/s}\).
The period is the length of the orbit divided by the speed, \(T = \dfrac{2\pi\,r}{v} = \dfrac{2\pi \cdot 6.8 \cdot 10^6}{7.69 \cdot 10^3}\) \(\approx 5.56 \cdot 10^3\ \text{s}\).
\(v \approx 7.7\ \text{km/s}\) and \(T \approx 93\ \text{min}\)
intermediate
Calculate the orbital radius of a geostationary satellite, with a period of 24 h, and its height above the ground. Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\), \(R_E = 6400\ \text{km}\) and \(\pi^2 \approx 9.87\).

Show solution
From \(T = 2\pi\sqrt{r^3/(G\,M)}\) we get \(r^3 = \dfrac{G\,M\,T^2}{4\pi^2}\), with \(T = 86\,400\ \text{s}\).
\(r^3 = \dfrac{4.02 \cdot 10^{14} \cdot (8.64 \cdot 10^4)^2}{4 \cdot 9.87}\) \(\approx 7.6 \cdot 10^{22}\ \text{m}^3\), and \(r = \sqrt[3]{7.6 \cdot 10^{22}}\).
\(r \approx 4.24 \cdot 10^7\ \text{m}\), about 42,000 km from the centre, or 36,000 km above the ground.
intermediate
How fast does a geostationary satellite move, given that its orbit has a radius of about 42,400 km?

Show solution
It goes once round in 24 h, and \(v = \dfrac{2\pi\,r}{T} = \dfrac{2\pi \cdot 4.24 \cdot 10^7}{86\,400}\).
\(v \approx 3.1\ \text{km/s}\)
intermediate
A satellite moves to a circular orbit with a radius four times larger. What happens to its speed and to its period?

Show solution
Since \(v = \sqrt{G\,M/r}\), quadrupling \(r\) divides \(v\) by \(\sqrt 4 = 2\).
Since \(T \propto r^{3/2}\), the period is multiplied by \(4^{3/2} = 8\).
The speed falls to half and the period becomes 8 times longer.
intermediate
Why are all geostationary satellites above the equator? Would a satellite with a 24 h period in an inclined orbit also stay still in the sky?

Show solution
Every orbit has the centre of the Earth in its plane, and the satellite needs to keep up with the point on the ground turning below it.
Only points on the equator turn in a circle centred on the centre of the Earth, and only there can the plane of the orbit coincide with that of the ground's motion.
In an inclined 24 h orbit, the satellite would spend half the day north of the equator and half south, tracing a figure of eight in the sky.
Only an equatorial 24 h orbit, in the direction of rotation, keeps the satellite still in the sky.
challenge
How many times does the ISS go round the Earth in a day? Use the period of about 92.6 min we calculated for the orbit of radius 6800 km.

Show solution
A day has 1440 min, and the number of laps is \(\dfrac{1440}{92.6}\).
About 15.5 laps a day, and the astronauts see the Sun rise some 15 or 16 times every 24 h.
challenge
Newton's cannon stands on top of an imaginary mountain, 6800 km from the centre of the Earth, above the air. The ball leaves horizontally at 9 km/s, faster than the 7.7 km/s of the circular orbit at that height. Describe the path and calculate the greatest distance from the centre of the Earth. Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and the fact that, in an elliptical orbit, the energy per kilogram is \(-G\,M/(2a)\), where \(a\) is the semi-major axis.

Show solution
The energy per kilogram at firing is \(\dfrac{v^2}{2} - \dfrac{G\,M}{r}\) \(= \dfrac{(9 \cdot 10^3)^2}{2} - \dfrac{4.02 \cdot 10^{14}}{6.8 \cdot 10^6}\) \(\approx 4.05 \cdot 10^7 - 5.91 \cdot 10^7\) \(= -1.86 \cdot 10^7\ \text{J/kg}\).
Being negative, it indicates a closed orbit, an ellipse, with \(a = \dfrac{G\,M}{2 \cdot 1.86 \cdot 10^7}\) \(\approx 1.08 \cdot 10^7\ \text{m}\).
The firing point is the perigee, and the furthest point is at \(r_a = 2a - r_p\) \(= 2.16 \cdot 10^7 - 6.8 \cdot 10^6\).
The ball follows an ellipse and reaches about \(1.48 \cdot 10^7\ \text{m}\) from the centre, some 8400 km above the ground, returning to the top of the mountain on every lap.

STEP 4

Do astronauts float because there is no gravity?

The images from the International Space Station, with astronauts and objects floating around, suggest a place without gravity. We saw in step 2, however, that at a height of 400 km \(g\) is still about 8.7 m/s², and it is this gravity that keeps the station in orbit.

The clue, as we shall see, lies in the scales. They measure the force with which a person pushes on the floor, and this reading is called the apparent weight. When the lift accelerates, the scales have to exert a force different from the weight for the person to keep up with the lift, and the reading changes, even though gravity stays the same.

In a lift in free fall, the person, the scales and the lift all fall with the same acceleration \(g\), and nothing pushes on anything. The reading goes to zero, and it feels like floating. The space station is in this same situation, falling all the time round the Earth, and this state is called weightlessness.

Parabolic-flight aircraft, used in astronaut training, reproduce this for about 20 seconds at a time. The pilot makes the plane follow the same parabola a thrown stone would follow, and inside it everything floats.

\(N - m\,g = m\,a\)\(N = m\,(g + a)\)\(a = -g \Rightarrow N = 0\)\(N\) is the force of the scales on the person, and \(a\) is the acceleration of the lift, positive upwards. Like bathroom scales, the ones in the simulation show \(N/9.8\) in kilograms, and the lift accelerates at 2 m/s² when it is not in free fall.

Let's discuss

Compare the two bars with the lift at rest and then accelerating upwards. Which of them changed?
Release the ball with the lift at rest and then in free fall. What does the ball seem to do, relative to the lift, in each case?
Choose the station in orbit. Has the gravity bar gone to zero? If not, how can we explain the scales reading zero?
Change the person's mass during the parabolic flight. Does the scales reading depend on the mass?

Person's mass

Local g

Cabin acceleration

Force of gravity

Scales reading

Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
A 60 kg person is standing on scales in a lift at rest. What do the scales read, in newtons? Use \(g = 10\ \text{m/s}^2\).

Show solution
With the lift at rest, the acceleration is zero and \(N = m\,g = 60 \cdot 10\).
\(N = 600\ \text{N}\)
basic
The same lift goes up, accelerating at 2 m/s². What do the scales read now?

Show solution
The resultant force has to point upwards, \(N - m\,g = m\,a\), and so \(N = m\,(g + a) = 60 \cdot 12\).
\(N = 720\ \text{N}\), and the person feels heavier.
basic
The lift cable snaps and the lift falls freely. What do the scales read during the fall, if we neglect the air?

Show solution
In free fall, \(a = -g\), and \(N = m\,(g + a) = 60 \cdot (10 - 10)\).
The scales read zero, because the person and the scales fall together.
basic
Lucas claims that the astronauts on the ISS float because, at a height of 400 km, there is no longer any gravity. What is wrong with this explanation?

Show solution
At a height of 400 km, \(g\) is still about 8.7 m/s², almost 90% of the value on the ground, and it is this gravity that keeps the station in orbit.
The astronauts float because they and the station are in free fall together, falling round the Earth with the same acceleration.
Gravity is there, and strong; what is missing is a contact force, since nothing pushes the astronaut against the floor.
intermediate
The lift goes down, accelerating at 3 m/s². What do the scales read for the 60 kg person, in newtons, and what do scales that divide the force by 10 m/s² to give kilograms show?

Show solution
With the acceleration downwards, \(a = -3\ \text{m/s}^2\), and \(N = m\,(g + a) = 60 \cdot (10 - 3)\).
The scales in kilograms divide this force by 10.
\(N = 420\ \text{N}\), and the scales show 42 kg.
intermediate
In a parabolic flight, the plane is climbing with a vertical velocity of 100 m/s when the pilot starts adjusting the engines only to make up for the air, and the plane continues in free fall until the vertical velocity is 100 m/s downwards. How long does the weightlessness last? Use \(g = 10\ \text{m/s}^2\).

Show solution
In free fall, the vertical velocity changes from \(+100\) to \(-100\ \text{m/s}\), a change of 200 m/s, at a rate of 10 m/s every second, and \(\Delta t = \dfrac{200}{10}\).
\(\Delta t = 20\ \text{s}\) of weightlessness.
intermediate
The lift goes down at a constant speed of 5 m/s. What do the scales read for the 60 kg person?

Show solution
At constant velocity the acceleration is zero, and the scales only need to balance the weight, \(N = m\,g = 60 \cdot 10\).
\(N = 600\ \text{N}\), the same reading as with the lift at rest.
intermediate
Inside the ISS, the astronaut Helena lets go of a pen in front of her face. What does she see? And what would someone watching the scene from outside, at rest relative to the centre of the Earth, see?

Show solution
The pen, the astronaut and the station fall together round the Earth, with the same acceleration.
Relative to Helena, the pen stays still in the air, floating.
For someone watching from outside, the pen follows the same orbit as the station, at about 7.7 km/s, falling round the Earth.
Inside the station the pen floats, and from outside it orbits the Earth together with the station.
challenge
To create artificial gravity, a wheel-shaped space station spins about its own axis. With a radius of 50 m, what must the angular velocity be for an astronaut on the rim to feel an acceleration of 10 m/s²? How long does each turn take?

Show solution
The floor of the rim pushes the astronaut towards the centre and plays the part of the centripetal force, with \(a_c = \omega^2\,r\).
So \(\omega = \sqrt{a_c/r} = \sqrt{10/50}\), and the period is \(T = 2\pi/\omega\).
\(\omega \approx 0.45\ \text{rad/s}\), one turn every 14 s.
challenge
In a lift, the scales read 480 N for a 60 kg person. What is the acceleration of the lift? Can we tell whether it is going up or down? Use \(g = 10\ \text{m/s}^2\).

Show solution
From \(N = m\,(g + a)\) we get \(a = \dfrac{N}{m} - g = \dfrac{480}{60} - 10\).
A downward acceleration appears both in a lift going down faster and faster and in one going up and braking.
\(a = -2\ \text{m/s}^2\), or 2 m/s² downwards, and the scales alone do not tell us whether the lift is going up or down.

STEP 5

How do we weigh Jupiter without leaving the Earth?

In the Circular Motion lesson, in the step on Kepler's laws, we met the three laws Kepler drew from observations, among them the third, \(T^2/a^3 = \text{constant}\). We can now show where it comes from, because, by squaring the formula for the period of the circular orbit, we find \(T^2/r^3 = 4\pi^2/(G\,M)\).

The calculation says more than Kepler knew. The constant depends only on the mass \(M\) of the central body and holds for all the satellites that go round it, whatever the mass of each one. Jupiter's moons have one constant, the planets round the Sun have another and the Earth's artificial satellites, a third.

The result is an astronomical balance. Measuring the radius and the period of a single moon gives the mass of the planet, and this is how the mass of Jupiter was obtained, that of the Earth from the Moon, and that of the Sun from the Earth itself.

\(\dfrac{T^2}{r^3} = \dfrac{4\pi^2}{G\,M}\)\(M = \dfrac{4\pi^2\,r^3}{G\,T^2}\)\(r_p\,v_p = r_a\,v_a\)In elliptical orbits, \(r\) gives way to the semi-major axis \(a\). The last relation holds between perihelion (\(p\)) and aphelion (\(a\)), the points at which the velocity is perpendicular to the radius. The radii and periods in the simulation are approximate values, and the line is fitted to the points by the least-squares method.

The second law, the law of areas, also gets a new reading. It is equivalent to the conservation of angular momentum, and at perihelion and at aphelion the product of the distance and the speed is the same. The planet speeds up as it approaches the Sun, pulled forwards by gravity itself, and slows down as it moves away.

In the simulation, each satellite becomes a point on the graph of \(T^2\) against \(r^3\). If the theory is right, the points should fall, approximately, on a straight line through the origin, and the slope of this line, \(4\pi^2/(G\,M)\), hands us the central mass.

Let's discuss

With Jupiter's moons, check whether the four points lie on the same line. What mass does the simulation calculate for Jupiter?
Move the new body further out and further in. How many times does the period seem to increase when the radius doubles?
Switch to the Earth's satellites and then to the planets. Does the slope of the line change? Why?
For the Earth's satellites, pull the new body in as close as possible. Can we get to one lap an hour?

Radius of the new body

Calculated central mass

T²/r³

Radius of the new body

Period of the new body

Speed of the new body

Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Satellite B goes round the Earth in an orbit with a radius four times that of satellite A. How many times longer is B's period than A's?

Show solution
By the 3rd law, \(T^2 \propto r^3\), and \(\dfrac{T_B}{T_A} = \left(\dfrac{r_B}{r_A}\right)^{3/2} = 4^{3/2}\).
B's period is 8 times A's.
basic
Is the ratio \(T^2/r^3\) calculated for the Moon, round the Earth, equal to the one calculated for the Earth, round the Sun? Why?

Show solution
The ratio is \(\dfrac{4\pi^2}{G\,M}\), where \(M\) is the mass of the central body.
For the Moon, the central body is the Earth; for the Earth, it is the Sun, whose mass is about 330 thousand times greater.
The ratios are different, because each one depends on the mass of the body being orbited.
basic
The Earth passes closest to the Sun at the beginning of January, at perihelion, and furthest away at the beginning of July, at aphelion. In which of the two months does it move faster in its orbit?

Show solution
By the 2nd law, the Sun–Earth line sweeps out equal areas in equal times, which is equivalent to constant \(r\,v\) at these two points.
With a smaller \(r\), the speed has to be greater.
In January, at perihelion.
basic
Mars orbits the Sun at 1.52 AU. How long is the Martian year, in Earth years? Use \(T^2 = r^3\), with \(T\) in years and \(r\) in AU.

Show solution
\(T = \sqrt{r^3} = \sqrt{1.52^3} = \sqrt{3.51}\).
\(T \approx 1.87\) Earth years, about 1 year and 10 months.
intermediate
The Moon goes once round the Earth in 27.3 days, in an orbit of radius \(3.84 \cdot 10^8\ \text{m}\). Calculate the mass of the Earth. Use \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\) and \(\pi^2 \approx 9.87\).

Show solution
The period in seconds is \(T = 27.3 \cdot 86\,400 \approx 2.36 \cdot 10^6\ \text{s}\).
\(M = \dfrac{4\pi^2\,r^3}{G\,T^2}\) \(= \dfrac{4 \cdot 9.87 \cdot (3.84 \cdot 10^8)^3}{6.7 \cdot 10^{-11} \cdot (2.36 \cdot 10^6)^2}\).
\(M \approx 6.0 \cdot 10^{24}\ \text{kg}\), the same value we obtained from \(g\) and \(R_E\) in step 2.
intermediate
The Earth goes round the Sun at \(1.5 \cdot 10^{11}\ \text{m}\), with a period of one year, \(3.15 \cdot 10^7\ \text{s}\). Calculate the mass of the Sun, with \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\) and \(\pi^2 \approx 9.87\).

Show solution
\(M_S = \dfrac{4\pi^2\,r^3}{G\,T^2}\) \(= \dfrac{4 \cdot 9.87 \cdot (1.5 \cdot 10^{11})^3}{6.7 \cdot 10^{-11} \cdot (3.15 \cdot 10^7)^2}\).
\(M_S \approx 2.0 \cdot 10^{30}\ \text{kg}\)
intermediate
At perihelion, the Earth is 147 million km from the Sun; at aphelion, it is 152 million km away and moves at 29.3 km/s. What is its speed at perihelion?

Show solution
At both points the velocity is perpendicular to the radius, and Kepler's 2nd law gives \(r_p\,v_p = r_a\,v_a\), or \(v_p = \dfrac{152 \cdot 29.3}{147}\).
\(v_p \approx 30.3\ \text{km/s}\)
intermediate
Io, one of Jupiter's moons, has an orbit of radius \(4.22 \cdot 10^8\ \text{m}\) and a period of 1.77 days. Calculate the mass of Jupiter, with \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\) and \(\pi^2 \approx 9.87\).

Show solution
In seconds, \(T = 1.77 \cdot 86\,400 \approx 1.53 \cdot 10^5\ \text{s}\).
\(M_J = \dfrac{4\pi^2\,r^3}{G\,T^2}\) \(= \dfrac{4 \cdot 9.87 \cdot (4.22 \cdot 10^8)^3}{6.7 \cdot 10^{-11} \cdot (1.53 \cdot 10^5)^2}\).
\(M_J \approx 1.9 \cdot 10^{27}\ \text{kg}\), more than 300 times the mass of the Earth.
challenge
Is it possible for a satellite to go round the Earth once every hour? Calculate the shortest possible period, that of an orbit skimming the surface, and draw a conclusion. Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and \(R_E = 6.4 \cdot 10^6\ \text{m}\).

Show solution
\(T = 2\pi\sqrt{\dfrac{R_E^3}{G\,M_E}}\) \(= 2\pi\sqrt{\dfrac{(6.4 \cdot 10^6)^3}{4.02 \cdot 10^{14}}}\) \(\approx 5.07 \cdot 10^3\ \text{s}\).
A shorter period would require a radius smaller than the Earth's, because \(T\) grows with \(r\).
It is not possible, because no satellite of the Earth goes round in less than about 85 minutes.
challenge
Halley's comet returns every 76 years and, at perihelion, passes 0.59 AU from the Sun. Calculate the semi-major axis of its orbit and the aphelion distance. Use \(T^2 = a^3\), with \(T\) in years and \(a\) in AU, and remember that the major axis measures \(r_p + r_a = 2a\).

Show solution
The semi-major axis is \(a = T^{2/3} = 76^{2/3} = \sqrt[3]{5776}\).
The aphelion is at \(r_a = 2a - r_p \approx 35.9 - 0.59\).
\(a \approx 17.9\ \text{AU}\), and the aphelion is about 35 AU away, beyond the orbit of Neptune.

STEP 6

How fast do we have to go to escape from the Earth?

When a body rises, it trades kinetic energy for gravitational potential energy. Near the ground we used \(E_p = m\,g\,h\), but this formula assumes a constant \(g\), and for heights comparable with the Earth's radius we need the general expression, \(E_p = -G\,M\,m/r\).

The minus sign may be bothersome at first sight. We choose the zero of potential energy at infinity, where the attraction vanishes, and, since gravity pulls inwards, any point closer to the planet has less energy than there. Hence the negative value, which approaches zero as \(r\) increases. The graph of this energy looks like a well, and that is the picture we shall work with.

If the total energy \(E_k + E_p\) is negative, the body rises to a turning point and falls back. If it is zero or positive, the body goes away for ever. The smallest launch speed that brings the total energy to zero is the escape velocity, which on Earth is about 11.2 km/s, if we neglect the air.

\(E_p = -\dfrac{G\,M\,m}{r}\)\(\dfrac{m\,v_e^2}{2} - \dfrac{G\,M\,m}{R} = 0\)\(v_e = \sqrt{\dfrac{2\,G\,M}{R}}\)\(R\) is the radius of the planet, and the escape velocity does not depend on the mass of the projectile. In the simulation, we neglect the air and the rotation of the planet, the left-hand column goes up to 6 radii and the axis of the well, up to 10.

On the Moon, the escape velocity is only 2.4 km/s. The molecules of a gas jiggle around with speeds that, on average, stay below this, but some of them are much faster, and over billions of years these molecules would have escaped. This is probably the main reason the Moon has almost no atmosphere.

If we squeezed a mass into a small enough radius, the escape velocity would exceed the speed of light, and not even light would get out. This is the idea of a black hole, which Einstein's theory describes more carefully than our calculation does.

Let's discuss

On Earth, launch at 5 km/s and at 8 km/s. Where does the total-energy line touch the well, and at what height does the projectile stop?
Press the escape-velocity button. What happens to the total-energy line relative to zero?
Compare the escape velocities of the Earth, the Moon and Mars. Which of these bodies holds on to an atmosphere best?
At 8 km/s, the projectile rises some 6700 km. What would \(v^2/(2g)\) give, with constant \(g\), and how can we explain the difference?

Launch speed

Escape velocity

Total energy per kg

Predicted maximum height

Height now

Speed now

Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
Calculate the Earth's escape velocity with \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and \(R_E = 6.4 \cdot 10^6\ \text{m}\), neglecting the air.

Show solution
\(v_e = \sqrt{\dfrac{2\,G\,M_E}{R_E}}\) \(= \sqrt{\dfrac{2 \cdot 4.02 \cdot 10^{14}}{6.4 \cdot 10^6}}\) \(= \sqrt{1.26 \cdot 10^8}\).
\(v_e \approx 11.2\ \text{km/s}\)
basic
Why is the gravitational potential energy \(E_p = -G\,M\,m/r\) negative? Does it increase or decrease when the body moves away from the Earth?

Show solution
We choose the zero of potential energy at infinity, where the attraction vanishes.
To take a body from near the Earth all the way there, we have to supply energy, and so any point at a finite distance has energy below zero.
It is negative because of the choice of zero at infinity, and it increases, approaching zero, as the body moves away.
basic
Would a lorry and a marble need the same speed to escape from the Earth? And the same energy?

Show solution
In the escape-velocity calculation, \(\dfrac{m\,v_e^2}{2} = \dfrac{G\,M\,m}{R}\), the mass \(m\) cancels, and \(v_e\) does not depend on it.
The energy needed, \(G\,M\,m/R\), is proportional to the mass.
The speed is the same, 11.2 km/s, and the lorry needs much more energy.
basic
A 1000 kg satellite is 12,800 km from the centre of the Earth. What is its gravitational potential energy? Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\).

Show solution
\(E_p = -\dfrac{G\,M_E\,m}{r}\) \(= -\dfrac{4.02 \cdot 10^{14} \cdot 1000}{1.28 \cdot 10^7}\).
\(E_p \approx -3.1 \cdot 10^{10}\ \text{J}\)
intermediate
The Moon has a mass of \(7.4 \cdot 10^{22}\ \text{kg}\) and a radius of 1740 km. Calculate the Moon's escape velocity, with \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\).

Show solution
\(v_e = \sqrt{\dfrac{2 \cdot 6.7 \cdot 10^{-11} \cdot 7.4 \cdot 10^{22}}{1.74 \cdot 10^6}}\) \(= \sqrt{5.7 \cdot 10^6}\).
\(v_e \approx 2.4\ \text{km/s}\), a little over a fifth of the Earth's escape velocity.
intermediate
How much energy does it take to lift a 1000 kg satellite from the Earth's surface to a height of 400 km, not counting the kinetic energy of the orbit? Compare with \(m\,g\,h\), using \(g = 9.8\ \text{m/s}^2\). Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\).

Show solution
The change in potential energy is \(\Delta E_p = G\,M\,m\left(\dfrac{1}{R_E} - \dfrac{1}{r}\right)\), and the bracket is \(\dfrac{1}{6.4 \cdot 10^6} - \dfrac{1}{6.8 \cdot 10^6}\) \(\approx 9.2 \cdot 10^{-9}\ \text{m}^{-1}\), so that \(\Delta E_p = 4.02 \cdot 10^{17} \cdot 9.2 \cdot 10^{-9}\).
With constant \(g\), \(m\,g\,h = 1000 \cdot 9.8 \cdot 4 \cdot 10^5\) \(\approx 3.9 \cdot 10^9\ \text{J}\), slightly more, because in reality \(g\) decreases with height.
About \(3.7 \cdot 10^9\ \text{J}\), some 6% less than \(m\,g\,h\).
intermediate
A projectile is launched vertically from the Earth's surface at 5 km/s, with no air. How high does it rise? Compare with \(v^2/(2g)\), which assumes a constant \(g = 9.8\ \text{m/s}^2\). Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and \(R_E = 6.4 \cdot 10^6\ \text{m}\).

Show solution
By conservation of energy, \(\dfrac{v^2}{2} - \dfrac{G\,M}{R_E} = -\dfrac{G\,M}{r_{\text{max}}}\), and \(\dfrac{G\,M}{r_{\text{max}}} = 6.28 \cdot 10^7 - 1.25 \cdot 10^7\) \(= 5.03 \cdot 10^7\ \text{J/kg}\).
So \(r_{\text{max}} = \dfrac{4.02 \cdot 10^{14}}{5.03 \cdot 10^7} \approx 7.99 \cdot 10^6\ \text{m}\), and the height is \(r_{\text{max}} - R_E\).
With constant \(g\), it would give \(\dfrac{(5 \cdot 10^3)^2}{2 \cdot 9.8} \approx 1.28 \cdot 10^6\ \text{m}\).
About 1590 km, well above the 1280 km from the constant-\(g\) calculation, because gravity weakens during the climb.
intermediate
The Earth has an atmosphere and the Moon practically none. Explain the difference using escape velocity.

Show solution
The molecules of a gas jiggle about in all directions, and a fraction of them, however small, moves much faster than the average.
On the Moon, the escape velocity is only 2.4 km/s, and over billions of years these faster molecules kept escaping, while the Earth, with 11.2 km/s, holds on to most of its air.
The Moon's weak gravity cannot hold a gas for billions of years, and whatever atmosphere it may have had has escaped.
challenge
Down to what radius would we have to squeeze the Earth, keeping its mass, for the escape velocity at the surface to equal the speed of light, \(3 \cdot 10^8\ \text{m/s}\)? Use the Newtonian formula, which in this case coincides with the relativistic one, and \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\).

Show solution
Setting \(v_e = c\) in \(v_e = \sqrt{2\,G\,M/R}\), we have \(R = \dfrac{2\,G\,M}{c^2}\) \(= \dfrac{2 \cdot 4.02 \cdot 10^{14}}{9 \cdot 10^{16}}\).
\(R \approx 9 \cdot 10^{-3}\ \text{m}\), less than 1 cm, the size of a marble.
challenge
A satellite of mass \(m\) is in a circular orbit of radius \(r\). Show that its kinetic energy is \(G\,M\,m/(2r)\) and its total mechanical energy is \(-G\,M\,m/(2r)\). How much more speed would an object in the ISS orbit, with \(r = 6800\ \text{km}\) and \(v = 7.7\ \text{km/s}\), need to escape from the Earth?

Show solution
From the circular orbit, \(m\,v^2 = G\,M\,m/r\), and so \(E_k = \dfrac{m\,v^2}{2} = \dfrac{G\,M\,m}{2r}\).
Adding the potential energy, \(E = \dfrac{G\,M\,m}{2r} - \dfrac{G\,M\,m}{r}\) \(= -\dfrac{G\,M\,m}{2r}\).
To escape from there, the energy has to reach zero, which calls for \(v_e = \sqrt{2}\,v\) \(\approx 1.41 \cdot 7.7 \approx 10.9\ \text{km/s}\).
About 3.2 km/s more, added in the direction of motion.

STEP 7

Why does the sea rise twice a day?

Anyone who spends a day at the beach sees the sea rise and fall, and fishermen know that there are, as a rule, two high tides a day, a little over 12 hours apart. The main cause is the Moon, and what matters here is that its pull has different strengths at different points of the Earth.

The side of the Earth facing the Moon is closer to it and is pulled harder than the centre, and the centre is pulled harder than the far side. Seen from the centre of the Earth, it is as if the ocean were stretched in both directions, forming two bulges, one facing the Moon and the other on the opposite side. The Earth turns underneath them, and each port goes through two high tides on every turn.

This difference is called the tidal force. It falls with the cube of the distance, faster than gravity itself, and this is why the Sun, which pulls the Earth with a force about 180 times greater than the Moon's, produces tides less than half as strong as the lunar ones.

At new moon and full moon, the Sun, the Earth and the Moon are lined up, the two effects add together, giving the spring tides, with the largest range. At first and last quarter, the Sun and the Moon pull in perpendicular directions, and the neap tides are weaker.

\(a_{\text{tide}} \approx \dfrac{2\,G\,M_M\,R_E}{d^3}\)\(\dfrac{\text{Sun's tide}}{\text{Moon's tide}} \approx 0.45\)\(M_M = 7.4 \cdot 10^{22}\ \text{kg}\) is the mass of the Moon and \(d\) is the distance between the centres. In the simulation, the deformation of the ocean is greatly exaggerated and the distances are not to scale, and each day lasts 10 s. We ignore the continents and the friction of the water, which delay and distort the real tides.

tidal force, seen from the centre of the Earth

Let's discuss

Stop the Moon and follow the port for a day. How many high tides appear on the graph?
Set the Moon moving. Is the interval between two high tides still 12 h?
Go from new moon to first quarter. How does the difference between high and low tide change?
Look at the arrows on the side opposite the Moon. How can we explain that they point away from it?

Position of the Moon

Phase of the Moon

Type of tide

Range (Moon alone = 1)

Two high tides every

Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge

basic
How many high tides happen per day at a port, and roughly what is the interval between two of them?

Show solution
The Moon leaves the ocean with two bulges, one facing it and the other on the opposite side, and the Earth turns underneath both.
Since the Moon also moves along its orbit, the Earth needs about 24 h 50 min to return to the same position relative to it.
Two high tides a day, about 12 h 25 min apart.
basic
At which phases of the Moon do the spring tides, those with the largest range, happen, and why?

Show solution
At new moon and full moon, the Sun, the Earth and the Moon are lined up.
The bulges caused by the Sun and by the Moon point in the same direction and add together.
At new moon and at full moon, when the effects of the Sun and the Moon add together.
basic
And the neap tides, with the smallest range, at which phases do they happen?

Show solution
At first and last quarter, the direction of the Moon is perpendicular to that of the Sun, seen from the Earth.
The bulge caused by the Sun lies where the Moon produces low tide, and one cancels part of the other.
At first and last quarter.
basic
If the Moon pulls the water towards itself, why is there also a high tide on the side of the Earth opposite the Moon?

Show solution
The Moon pulls the centre of the Earth harder than the water on the far side, which is further away.
Seen from the centre of the Earth, it is as if this water were left behind, moving away from the Moon.
The high tide on the far side appears because the Earth is pulled away from the water on that side, and what counts is the difference between the pulls.
intermediate
Calculate the difference between the acceleration the Moon causes at the point of the Earth closest to it and at the centre of the Earth, using the approximation \(a_{\text{tide}} \approx 2\,G\,M_M\,R_E/d^3\). Data: \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\), \(M_M = 7.4 \cdot 10^{22}\ \text{kg}\), \(R_E = 6.4 \cdot 10^6\ \text{m}\) and \(d = 3.84 \cdot 10^8\ \text{m}\).

Show solution
The numerator is \(2 \cdot 6.7 \cdot 10^{-11} \cdot 7.4 \cdot 10^{22} \cdot 6.4 \cdot 10^6\) \(\approx 6.35 \cdot 10^{19}\), and the denominator is \((3.84 \cdot 10^8)^3 \approx 5.66 \cdot 10^{25}\).
The tidal acceleration is the ratio between the two.
\(a_{\text{tide}} \approx 1.1 \cdot 10^{-6}\ \text{m/s}^2\), about a ten-millionth of \(g\).
intermediate
The high tides at a port happen, on average, some 50 minutes later each day. Explain this delay, given that the Moon goes once right round the sky, relative to the Sun, every 29.5 days.

Show solution
While the Earth turns once, the Moon moves along its orbit, in the same direction, by about \(\dfrac{360^\circ}{29.5} \approx 12.2^\circ\).
The Earth turns \(15^\circ\) per hour and needs an extra \(\dfrac{12.2}{15}\ \text{h}\) to catch up with the Moon.
About 0.81 h, or 49 min per day, because the Moon moves along its orbit while the Earth turns.
intermediate
The Sun pulls the Earth with a force about 180 times greater than the Moon's, and even so the solar tides are weaker than the lunar ones. Calculate the ratio between the tidal effects of the Sun and of the Moon, which depends on \(M/d^3\). Use \(M_S = 2 \cdot 10^{30}\ \text{kg}\), \(d_S = 1.5 \cdot 10^{11}\ \text{m}\), \(M_M = 7.4 \cdot 10^{22}\ \text{kg}\) and \(d_M = 3.84 \cdot 10^8\ \text{m}\).

Show solution
The ratio is \(\dfrac{M_S}{M_M}\left(\dfrac{d_M}{d_S}\right)^3\) \(= \dfrac{2 \cdot 10^{30}}{7.4 \cdot 10^{22}} \cdot \left(\dfrac{3.84 \cdot 10^8}{1.5 \cdot 10^{11}}\right)^3\) \(\approx 2.7 \cdot 10^7 \cdot 1.68 \cdot 10^{-8}\).
The Sun's tidal effect is about 0.45 of the Moon's, less than half, because it depends on the cube of the distance.
intermediate
With the Sun's tidal effect equal to 0.45 of the Moon's, estimate the ratio between the range of the spring tides and that of the neap tides.

Show solution
At spring tide the effects add, \(1 + 0.45 = 1.45\), and at neap tide one subtracts from the other, \(1 - 0.45 = 0.55\).
The ratio is \(\dfrac{1.45}{0.55}\).
Spring tides have a range about 2.6 times that of neap tides.
challenge
If the Moon were at half its present distance, how many times stronger would the lunar tides be? And how many times greater would the force with which the Moon pulls the Earth be?

Show solution
The tidal effect is proportional to \(1/d^3\), and at half the distance it becomes \(2^3\) times greater.
The gravitational force is proportional to \(1/d^2\) and becomes \(2^2\) times greater.
Tides 8 times stronger and a force 4 times greater.
challenge
The Earth also raises tides on the Moon. Calculate the Earth's tidal effect on the Moon, \(2\,G\,M_E\,R_M/d^3\), with \(R_M = 1.74 \cdot 10^6\ \text{m}\), and compare it with that of the Moon on the Earth, \(1.1 \cdot 10^{-6}\ \text{m/s}^2\). Use \(G\,M_E = 4.02 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and \(d = 3.84 \cdot 10^8\ \text{m}\).

Show solution
\(a = \dfrac{2 \cdot 4.02 \cdot 10^{14} \cdot 1.74 \cdot 10^6}{(3.84 \cdot 10^8)^3}\) \(= \dfrac{1.40 \cdot 10^{21}}{5.66 \cdot 10^{25}}\) \(\approx 2.5 \cdot 10^{-5}\ \text{m/s}^2\).
Dividing by the Moon's effect on the Earth, \(\dfrac{2.5 \cdot 10^{-5}}{1.1 \cdot 10^{-6}}\).
The tides the Earth raises on the Moon are about 22 times stronger, and over millions of years this strain slowed the Moon's rotation, so that today it always shows us the same face.

WRAP-UP

Challenges

Law of gravitation

The Sun has a mass of \(2 \cdot 10^{30}\ \text{kg}\) and is \(1.5 \cdot 10^{11}\ \text{m}\) from the Earth; the Moon has \(7.4 \cdot 10^{22}\ \text{kg}\) and is \(3.84 \cdot 10^8\ \text{m}\) away. How many times harder does the Sun pull the Earth than the Moon does?

Show solution

The ratio between the forces is \(\dfrac{M_S}{M_M}\left(\dfrac{d_M}{d_S}\right)^2\) \(\approx 2.7 \cdot 10^7 \cdot 6.55 \cdot 10^{-6}\).
The Sun pulls the Earth with a force about 180 times greater than the Moon's, and it is round the Sun that the Earth travels.

Gravitational field

At what altitude is the weight of a body 1% less than at the Earth's surface? Use \(R_E = 6400\ \text{km}\).

Show solution

We want \(\left(\dfrac{R_E}{R_E + h}\right)^2 = 0.99\), and so \(\dfrac{R_E + h}{R_E} = \dfrac{1}{\sqrt{0.99}}\) \(\approx 1.005\).
The altitude is \(h \approx 0.005 \cdot 6400\), about 32 km, almost four times the height of Everest.

Orbits

GPS satellites go round twice a day, with a period of 12 h. If the geostationary orbit, of 24 h, has a radius of 42,400 km, what is the radius of the GPS orbit?

Show solution

By the 3rd law, \(r \propto T^{2/3}\), and \(r_{\text{GPS}} = 42\,400 \cdot \left(\dfrac{1}{2}\right)^{2/3}\) \(\approx 42\,400 \cdot 0.63\).
The radius comes out close to 26,700 km, some 20,000 km above the ground.

Weightlessness

A 70 kg astronaut is on the ISS, where \(g \approx 8.7\ \text{m/s}^2\). What is the gravitational force on her, and what do scales fixed to the floor of the station read?

Show solution

The force of gravity is \(W = m\,g = 70 \cdot 8.7\), about 610 N.
The scales read zero, because the astronaut and the station fall together round the Earth and nothing pushes one against the other.

Kepler's laws

If the Moon went round the Earth at half its present distance, what would its period be? Today it is 27.3 days.

Show solution

By the 3rd law, \(T \propto r^{3/2}\), and \(T' = 27.3 \cdot \left(\dfrac{1}{2}\right)^{3/2}\) \(\approx 27.3 \cdot 0.354\).
The new period would be about 9.7 days.

Escape velocity

Mars has a mass of \(6.4 \cdot 10^{23}\ \text{kg}\) and a radius of 3400 km. What is the escape velocity of Mars?

Show solution

\(v_e = \sqrt{\dfrac{2 \cdot 6.7 \cdot 10^{-11} \cdot 6.4 \cdot 10^{23}}{3.4 \cdot 10^6}}\) \(= \sqrt{2.52 \cdot 10^7}\).
The escape velocity comes out close to 5.0 km/s, less than half the Earth's, and a probe returning from Mars uses much less fuel to leave it than it used to leave here.

Tides

The Sun pulls the Earth about 180 times harder than the Moon does and, even so, produces smaller tides. How can we explain this?

Show solution

The tide depends on the difference between the pull on one side of the Earth and the pull on the other, and this difference falls with \(1/d^3\).
The Sun is about 390 times further away than the Moon, and the cube of this distance reduces its effect to some 45% of the lunar effect.

Think it through, no maths

If the Earth shrank to half its radius, without losing mass, what would happen to the Moon's orbit? And to our weight, on the new surface?

Show solution

The Moon would notice nothing, because the force on it depends only on the mass of the Earth and on the distance between the centres, which do not change.
On the new surface, we would be at half the distance to the centre, and our weight would become four times greater.

ENEM-style questions

The ENEM is Brazil's national secondary-school exam, which most students sit to get into university, and we wrote the five questions below in its format, with a base text taken from a real situation, a question and five options, each one returning to a different step of the lesson. Further down are real exam questions on gravitation, which may well be the best practice once these are done.

Gravitational field · Step 2
A space agency is testing a 300 kg exploration robot. Its legs can withstand, without bending, a force of at most 1200 N against the ground, and the team wants to know on which bodies it could stand without damage. The table gives approximate values of \(g\) at the surface.

g at the surface
Body g (m/s²)

Moon 1.6

Mercury 3.7

Mars 3.7

Venus 8.9

Earth 9.8

Assuming the robot is standing still on the ground, on which bodies can its legs bear its weight?
1. Only on the Moon.
2. On the Moon, on Mercury and on Mars.
3. On all the bodies in the table, because the robot's mass is the same everywhere.
4. Only on Earth, where the robot was designed and tested.
5. On none of them, because a mass of 300 kg always corresponds to 3000 N.
Show solution

Answer: B.
The weight is \(W = m\,g\), and the legs hold as long as \(300\,g \le 1200\), that is, as long as \(g \le 4\ \text{m/s}^2\).
On the Moon the weight is 480 N, and on Mercury and on Mars it is \(300 \cdot 3.7 = 1110\ \text{N}\), below the limit. On Venus and on Earth it exceeds 2600 N.
Option C confuses mass with weight, and E uses \(g = 10\ \text{m/s}^2\) on every body. Option A forgets Mercury and Mars, which are close to the limit, and D does no calculation at all.
Orbits · Step 3
At Teodoro's house, the satellite dish always points at the same spot in the sky, with no motor to follow the satellite. The technician explained that the TV satellite is geostationary, that is, it orbits above the equator and goes round once every 24 h, keeping pace with the Earth's rotation. Let us assume the orbit is circular and use \(G\,M_E = 4.0 \cdot 10^{14}\ \text{m}^3/\text{s}^2\) and \(\pi^2 \approx 10\).

What, approximately, is the radius of this satellite's orbit, measured from the centre of the Earth?
1. 180 km
2. 6400 km
3. 36,000 km
4. 42,000 km
5. 380,000 km
Show solution

Answer: D.
Setting gravity equal to the centripetal force, \(T = 2\pi\sqrt{r^3/(G\,M)}\), and so \(r^3 = \dfrac{G\,M\,T^2}{4\pi^2}\), with \(T = 86\,400\ \text{s}\).
\(r^3 = \dfrac{4.0 \cdot 10^{14} \cdot (8.64 \cdot 10^4)^2}{40}\) \(\approx 7.5 \cdot 10^{22}\ \text{m}^3\), and \(r \approx 4.2 \cdot 10^7\ \text{m}\).
Option C is the height above the ground, not the radius measured from the centre. Option A appears when we leave the period in hours, B is the radius of the Earth itself and E, the distance to the Moon.
Weightlessness · Step 4
A news report on the International Space Station says that the astronauts float because, at a height of 400 km, they are 'out of reach of the Earth's gravity'. The station goes once round the Earth every 93 minutes, and at that height the value of \(g\) is about 8.7 m/s².

Which statement explains why the astronauts float inside the station?
1. Gravity at a height of 400 km is zero, as the report says.
2. The Moon's pull balances the Earth's at that height.
3. The station and the astronauts are in free fall together round the Earth, and nothing pushes the astronauts against the floor.
4. A centrifugal force cancels gravity, and the resultant force on the astronauts is zero.
5. The lack of air in space removes the weight of bodies.
Show solution

Answer: C.
With \(g \approx 8.7\ \text{m/s}^2\), gravity still has almost 90% of its value on the ground, and it is gravity that makes the station go round the Earth, as the centripetal force.
Astronauts and station fall with the same acceleration, and the scales, or the floor, do not need to push anyone.
Option A repeats the report's mistake, and B greatly overestimates the Moon's pull. Option D forgets that the resultant force on the astronauts is not zero, since they are moving along a curve, and E confuses the lack of air with the lack of weight.
Escape velocity · Step 6
A company is studying launching small capsules of rock samples from the surface of the Moon, using an electromagnetic rail instead of rockets. The Moon has no air, and its mass and radius are \(7.4 \cdot 10^{22}\ \text{kg}\) and \(1.74 \cdot 10^6\ \text{m}\). Use \(G = 6.7 \cdot 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\).

What is the lowest speed at which the capsule must leave the rail to escape the Moon's pull?
1. 1.6 km/s
2. 1.7 km/s
3. 2.4 km/s
4. 11.2 km/s
5. 5700 km/s
Show solution

Answer: C.
The capsule escapes when the total energy reaches zero, \(\dfrac{m\,v_e^2}{2} = \dfrac{G\,M\,m}{R}\), and its mass cancels.
\(v_e = \sqrt{\dfrac{2 \cdot 6.7 \cdot 10^{-11} \cdot 7.4 \cdot 10^{22}}{1.74 \cdot 10^6}}\) \(\approx \sqrt{5.7 \cdot 10^6}\ \text{m/s}\), or about 2400 m/s.
Option B forgets the factor of 2 and gives the speed of a skimming orbit, and E forgets the square root. Option D is the Earth's escape velocity, and A mistakes the answer for the value of \(g\) on the Moon.

g at the surface
Body	g (m/s²)
Moon	1.6
Mercury	3.7
Mars	3.7
Venus	8.9
Earth	9.8

Tides · Step 7

In a fishing village, Marilene notes down the difference between high tide and low tide each day, in metres, together with the phase of the Moon. The table shows four records from the same month.

Records for the month
Day	Phase of the Moon	Difference (m)
1	new	2.6
8	first quarter	1.0
15	full	2.5
22	last quarter	1.1

Which explanation is consistent with these records?

Tides are caused only by the Sun, which is closer to the Earth at new moon and full moon.
At new moon and full moon, the Sun and the Moon are lined up with the Earth and their tidal effects add together; at the quarters, they act in perpendicular directions and one cancels part of the other.
At full moon the Moon is closer to the Earth and at new moon further away, and that is why the differences vary.
The larger tides happen when the Moon is lit up, because sunlight warms the water and makes it expand.
At first and last quarter, the Moon is on the other side of the Sun and its pull cancels out.

Show solution

Answer: B.
The large differences appear at new moon and full moon, the spring tides, and the small ones at the quarters, the neap tides.
If the Sun's effect is about 0.45 of the Moon's, the expected ratio is \(\dfrac{1 + 0.45}{1 - 0.45} \approx 2.6\), close to \(\dfrac{2.6}{1.0}\) in the table.
Options A and C invent distances that do not change in this way over the month, and D swaps gravity for heat. Option E gets the geometry wrong, because at the quarters the Moon makes a right angle with the Sun, seen from the Earth.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where each question can be looked up by year, day, booklet colour and number.

ENEM 2009, Day 1, blue booklet, question 27. In front of the Hubble telescope, in orbit, an astronaut remarks that it has a large mass and a small weight, and it is up to us to judge the statement by separating mass from weight and remembering the role of gravity in the orbit.
ENEM 2012, Day 1, blue booklet, question 74. A record of the positions of Mars in the sky, made every 10 days, shows a loop, which we need to explain by the difference between the orbital speeds of the Earth and of Mars.
ENEM 2018, Day 2, blue booklet, question 142. In the Mathematics paper, a graph places three satellites by their mass and orbital radius, and the law of gravitation lets us rank the forces the Earth exerts on them.
ENEM 2022, Day 2, blue booklet, question 123. The Sun is replaced, in a thought experiment, by a black hole of the same mass, and the question asks what would happen to the orbits of the planets.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Universal gravitation\(F = G\,M\,m/d^2\)

Lesson data\(G = 6.7 \cdot 10^{-11}\), \(M_E = 6 \cdot 10^{24}\ \text{kg}\), \(R_E = 6400\ \text{km}\)

Gravitational field\(g = G\,M/r^2\)

Weight\(W = m\,g\)

Scales in a lift\(N = m\,(g + a)\)

Circular orbit\(v = \sqrt{G\,M/r}\)

Orbital period\(T = 2\pi\sqrt{r^3/(G\,M)}\)

Kepler's 3rd law\(T^2/r^3 = 4\pi^2/(G\,M)\)

Kepler's 2nd law\(r_p\,v_p = r_a\,v_a\)

Potential energy\(E_p = -G\,M\,m/r\)

Escape velocity\(v_e = \sqrt{2\,G\,M/R}\)

Tidal force\(a_{\text{tide}} \approx 2\,G\,M\,R/d^3\)