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g from a planet's size and mass

Gravity and orbits · level 6, lesson 2 of 6

A question to start

Mars has about a tenth of the Earth's mass. So why is its gravity not a tenth of ours, but 38%?

The idea

On the surface of a planet of mass \(M\) and radius \(R\), the pull on a mass \(m\) is \(GMm/R^2\). That pull is the weight \(mg\). So

\[mg = \frac{GMm}{R^2} \qquad\Rightarrow\qquad g = \frac{GM}{R^2}.\]

The \(m\) cancels: every mass falls with the same \(g\) (level 1's heavy and light balls, now proved).

Mars. \(GM = 4.2828 \times 10^{13}\) m³/s² (measured from spacecraft orbits, more precisely than \(G\) or \(M\) alone). With the mean radius 3,389.5 km:

\[g = \frac{4.2828\times10^{13}}{(3.3895\times10^6)^2} = \frac{4.2828\times10^{13}}{1.1489\times10^{13}} = 3.73\ \text{m/s}^2.\]

At the equator, where the radius is 3,396.2 km, it is 3.71 m/s², the figure NASA gives. (The spin of Mars takes off another 0.017 m/s² there, which a scale would also feel.)

Comparing by ratios. For Mars against Earth,

\[\frac{g_M}{g_E} = \frac{M_M/M_E}{(R_M/R_E)^2} = \frac{0.107}{0.532^2} = \frac{0.107}{0.283} = 0.38.\]

So a tenth of the mass, but standing about half as far from the middle, which on its own would make gravity four times stronger. A tenth times about three and a half gives 0.38.

With density. Write \(M = \tfrac43\pi R^3 \rho\), where \(\rho\) is the mean density. Then \(g = \tfrac43 \pi G \rho R\): gravity grows with size and with density. Mars's rock is less dense (3,930 kg/m³ against 5,510 for Earth), partly because its iron core is smaller.

Mars Earth
Mass \(6.42 \times 10^{23}\) kg \(5.97 \times 10^{24}\) kg
Mean radius 3,389.5 km 6,371 km
\(GM\) 42,828 km³/s² 398,600 km³/s²
Mean density 3,930 kg/m³ 5,510 kg/m³
Surface gravity \(g\) 3.71 m/s² 9.81 m/s²

Worked examples

1. The Moon. \(GM = 4.905 \times 10^{12}\) m³/s², \(R = 1{,}737\) km. \(g = 4.905\times10^{12} / (1.737\times10^6)^2 = 1.63\) m/s². ✓

2. Phobos. \(GM \approx 7.1 \times 10^5\) m³/s², mean radius about 11.1 km. \(g = 7.1\times10^5 / (1.11\times10^4)^2 \approx 0.0058\) m/s², about 1/1700 of Earth's.

3. A planet twice as big, same rock. \(g \propto \rho R\), so its gravity would be twice as strong.

Try it

  1. Venus has \(GM = 3.249 \times 10^{14}\) m³/s² and \(R = 6{,}052\) km. Find \(g\) on Venus.
  2. A planet has the Earth's mass but twice its radius. What is its \(g\)?
  3. Show that \(g = \tfrac43\pi G\rho R\).
  4. Use the formula in question 3 to find \(g\) for Mars from its density and radius.
  5. Why does \(m\) cancel, and what does that mean?
Answers
  1. \(3.249\times10^{14} / (6.052\times10^6)^2 = 3.249\times10^{14} / 3.663\times10^{13} = 8.87\) m/s².
  2. \(9.81 / 4 = 2.45\) m/s².
  3. \(g = GM/R^2 = G \cdot \tfrac43\pi R^3\rho / R^2 = \tfrac43\pi G\rho R\).
  4. \(\tfrac43 \pi \times 6.674\times10^{-11} \times 3930 \times 3.3895\times10^6 = 3.72\) m/s².
  5. Both the pull and the resistance to speeding up grow with \(m\). So every mass falls with the same acceleration.

Common mistake

Forgetting to square the radius, or forgetting to change km to m. With \(GM\) in m³/s², \(R\) must be in metres.

Where it leads