A-Level Physics gravity calculations are either about calculating a scalar quantity which is gravitational potential energy or they are about calculating gravitational force, which is a vector. A typical gravitational potential energy question might be as follows:

Use the following data to calculate the gravitational potential at the surface of the Moon. Mass of Earth = 81 × mass of Moon, radius of Earth = 3.7 × radius of Moon, gravitational potential at the surface of the Earth = –63 MJ per kg

## Gravitational Potential Energy Questions

Gravitational potential is a *scalar* which means it has a magnitude but no direction. This makes it easy to calculate. All we need to know is the size of all the masses in the neighbourhood and their distances. Then we add up the potential energy from each mass. If the distance of a mass \(M\) is \(r\) then its contribution to the gravitational potential is

where \(G\) is the gravitational constant which is *extremely small* as gravity is a very weak force. G is just \(6.672 \times 10^{-11}\) which is ten zeros after the decimal point before we get to 6: 0.00000000006672. Going back to the question, if we are working out the gravitational potential energy at the Earth’s surface it will be

where \(M_e\) is the Earth’s mass and \(r_e\) is the Earth’s radius. On the surface of the Moon the potential energy will be

\(\text{GPE}_m = – \frac{G M_m}{r_m}\)The question is asking us about the relationship between the gravitational potential energy on the Moon’s surface expressed in terms of the gravitational potential energy on the Earth’s surface. We know that

\(M_e = 81 M_m\)and that

\(r_e = 3.7 r_m\)Substituting these into the equation for the potential on the lunar surface

\( \begin{align*}\text{GPE}_m &= – \frac{G M_m}{r_m} \\

&= – \frac{G \frac{M_e}{81}}{\frac{r_e}{3.7}} \\

&= – \frac{3.7}{81} \frac{G M_e}{r_e} \\

&= \frac{3.7}{81} \text{GPE}_e \\

\end{align*}

\)

Use your previous answer to sketch the gravitational potential between the surface of the Earth and the surface of the moon.

From the previous answer, we know that the Earth sits in a deep potential well and the Moon sits in a much shallower well which is just 5% the depth of that of the Earth. The potential looks like this:

Use the graph above to calculate how much energy it would take to throw a 1 kg rock to the maximum potential point between the Earth and the Moon.

The potential difference is 3.86 – 1.28 megajoules, which is 2.6 megajoules.

What would be the kinetic energy of the rock when it hit the surface of the Earth?

The potential difference between the maximum energy point and the surface of the Earth is 62.55-1.28, which is 61.3 megajoules. That’s 24 times the energy it took to throw the rock to the tipping point (which is called the Lagrange point as described below). This is why a moon base would be a strategic threat to the Earth.

## Gravitational Force Questions

The gravitational force between two masses \(M\) and \(m\) is given by

\(F= -\frac{GMm}{r^2}\)Whereas gravitational potential energy scales as \(\frac{1}{r}\) the gravitational force scales as \(\frac{1}{r^2}\). That’s because gravitational potential energy is the amount of work done to bring a mass from an infinite distance to some point in space and when we integrate \(1/r^2\) we get a constant times \(1/r\).

### Lagrange Point

Find the point between the Earth and the Moon at which the gravitational forces cancel one another out. Is this closer to the Moon or the Earth?

There is a point between the Earth and the Moon where the gravitational forces balance exactly called the *Lagrange point*. As the Moon has a lower mass than the Earth the Lagrange point is much closer to the Moon than the Earth. We can find this point by equating the gravitational force from the Earth and the Moon. The gravitational force between two masses \(M\) and \(m\) is given by

This scales as the square of the inverse of the separation of the masses, which means the gravitational force dwindles rapidly with distance. If we imagine a line from the centre of the Earth to the centre of the Moon the two forces we equate are

\( \begin{align*}F_e &= F_m \\

\frac{G M_e m}{r_e^2} &= \frac{G M_m m}{r_m^2} \\

\frac{M_e}{r_e^2} &= \frac{M_m}{r_m^2} \\

\frac{r_m^2}{r_e^2} &= \frac{M_m}{M_e} \\

\frac{r_m}{r_e} &= \sqrt{\frac{M_m}{M_e}}

\end{align*}

\)

As the ratio of the Earth’s mass to the Moon’s mass is about 81 the square root of that is 9. This means that the Earth-Moon Lagrange point is about 9 times as far from the Earth as from the Moon or about 11% of the way from the centre of the Moon to the centre of the Earth. If you do this calculation properly you have to correct for the fact that the whole system is in motion which increases the distance very slightly from 11% to 15% of the way from the Moon to Earth.

### Minimum Orbital Period for an Earth Satellite

A satellite of mass \(m\) travels at angular speed \(\omega\) in a circular orbit at a height \(h\) above the surface of a planet of mass \(M\) and radius \(R\). Using these symbols, give an equation that relates the gravitational force on the satellite to the centripetal force.

We know that if we are in a circular orbit the acceleration is

\(a = \omega^2 r\)And the centripetal force due to gravity is

\(F = \frac{GMm}{r^2}\)However, in this example we have to remember that the satellite is orbiting at a distance of \(r = R+h\) i.e. the radius of the Earth plus the height above the surface of the Earth. Now we can equate mass times orbital acceleration with the gravitational force:

\( \begin{align*}F &= m a \\

\frac{GMm}{(R+h)^2} &= m \omega^2 (R+h) \\

\frac{GM}{(R+h)^2} &= \omega^2 (R+h)

\end{align*}

\)

Use the equation above to show that the orbital period, T, of the satellite is given by \(T^2=\frac{4 \pi^2 (R+h)^3}{GM}\)

We know that \(\omega = \frac{2 \pi}{T}\) so we can substitute for \(\omega\) in the equation above to get

\( \begin{align*}\frac{GM}{(R+h)^2} &= \left( \frac{2 \pi}{T} \right)^2 (R+h) \\

\frac{GM}{(R+h)^2} &= \left( \frac{4 \pi^2}{T^2} \right)^2 (R+h) \\

T^2 &= \frac{4 \pi^2 (R+h)^3}{GM} \\

\end{align*}

\)

Explain why the period of a satellite in orbit around the Earth cannot be less than 85

minutes. Your answer should include a calculation to justify this value. Mass of the Earth = \(6.00 \times 10^24\) kg, radius of the Earth = \(6.40 \times 10^6\) m.

The orbital height of the satellite must be positive and so the limiting case is when \(h=0\).

\( \begin{align*}T^2 &= \frac{4 \pi^2 (R+h)^3}{GM} \\

&= \frac{4 \pi^2 (6.40 \times 10^6)^3}{6.672 \times 10^{-11} \times 6.00 \times 10^24} \\

&= 2.6 \times 10^7 \\

T &= 5084

\end{align*}

\)

This means that the minimum orbital period when \(h=0\) is 5084 seconds, which is 85 minutes. As the height of the orbit increases so will the orbital period so 85 minutes is a minimum orbital period for a satellite.

### Geosynchronous Orbit

The mass of the Earth \(6.00 × 10^24\) kg and its mean radius is \(6.40 \times 10^6\) m. Show that the radius of a geosynchronous orbit must be \(4.23 \times 10^7\) m.

We know that the orbital period for a geosynchronous satellite must be one day and that is enough to work out the radius of its orbit. For the orbit to be stable the Earth’s gravity must match the satellite’s mass times acceleration required for a circular orbit which is \(\omega^2 r\). We can use the relationship \(\omega = 2 \pi / T\) to substitute for the angular frequency where \(T\) is the length of a day in seconds which is \(24 \times 60 \times 60 = 86400\) seconds.

\(\begin{align*}\frac{GMm}{r^2} &= m \omega^2 r \\

GM &= \left ( \frac{2 \pi}{T} \right )^2 r^3 \\

GM &= \frac{4 \pi^2 r^3}{T^2} \\

r^3 &= \frac{GM T^2}{4 \pi^2} \\

r &= \sqrt[3]{\frac{GM T^2}{4 \pi^2}}

\end{align*}

\)

Putting in the numbers:

\(\begin{align*}r &= \sqrt[3]{\frac{6.672 \times 10^{-11} \times 6.00 × 10^24 \times 86400^2}{4 \pi^2}} \\

&= 4.23 \times 10^7 \\

\end{align*}

\)

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