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Weighing planets

Gravity and orbits · level 6, lesson 6 of 6

A question to start

Nobody has put Mars on a scale. Yet NASA's fact sheet says its mass is \(6.417 \times 10^{23}\) kg. How could anyone know that?

The idea

Turn Kepler's third law round:

\[T^2 = \frac{4\pi^2 r^3}{GM} \qquad\Rightarrow\qquad GM = \frac{4\pi^2 r^3}{T^2}.\]

Measure the size and period of any orbit round a planet, and you have its \(GM\). Divide by \(G\) and you have its mass.

Mars from Phobos. When Asaph Hall found the moons of Mars in 1877, he also gave astronomers a way to weigh Mars accurately for the first time. With modern values, \(r = 9.376 \times 10^6\) m and \(T = 7.654\) h \(= 27{,}554\) s:

\[GM = \frac{39.478 \times (9.376\times10^6)^3}{27554^2} = \frac{39.478 \times 8.242\times10^{20}}{7.592\times10^8} = 4.286 \times 10^{13}\ \text{m}^3/\text{s}^2.\]

The precise value, from spacecraft tracking, is \(4.2828 \times 10^{13}\): within 0.1%. (Phobos's orbit is slightly stretched and pulled by the bulge of Mars, which accounts for the small gap.) Then

\[M = \frac{GM}{G} = \frac{4.2828\times10^{13}}{6.674\times10^{-11}} = 6.42 \times 10^{23}\ \text{kg}.\]

Mars without a moon. For a planet with no moons (Venus, Mercury), astronomers waited for spacecraft. A probe's orbit, or its path as it flies past, works exactly like a moon's. Today, orbiters round Mars have measured its \(GM\) to better than one part in a million, and mapped how its gravity changes from place to place.

Why \(GM\) is better known than \(M\). Orbits give \(GM\) directly. To get \(M\) you must divide by \(G\), which is only known to about 1 part in 50,000. So spacecraft engineers work with \(GM\).

Worked examples

1. The Sun from the Earth. \(r = 1.496 \times 10^{11}\) m, \(T = 3.156 \times 10^7\) s. \(GM_\odot = 39.478 \times 3.348\times10^{33} / 9.960\times10^{14} = 1.327 \times 10^{20}\) m³/s², and \(M_\odot = 1.99 \times 10^{30}\) kg.

2. Earth from the Moon. \(r = 3.844 \times 10^8\) m, \(T = 27.32\) days \(= 2.360 \times 10^6\) s. \(GM = 39.478 \times 5.680\times10^{25} / 5.570\times10^{12} = 4.03 \times 10^{14}\) m³/s². The true value is \(3.986 \times 10^{14}\); the 1% gap is because the Moon is heavy enough that the Earth and Moon both orbit their shared centre of mass (the formula then gives \(G(M + m)\)).

3. Mars from Deimos. \(r = 2.3463 \times 10^7\) m, \(T = 1.091 \times 10^5\) s. \(GM = 39.478 \times 1.2917\times10^{22} / 1.190\times10^{10} = 4.285 \times 10^{13}\). ✓

Try it

  1. Jupiter's moon Io orbits at 421,700 km in 1.769 days. Find Jupiter's \(GM\) and mass.
  2. A probe orbits an asteroid at 20 km in 10 hours. Find the asteroid's mass.
  3. Why does the Moon give a slightly wrong mass for the Earth, while Phobos gives a good mass for Mars?
  4. Why do engineers use \(GM\) rather than \(M\)?
Answers
  1. \(T = 152{,}840\) s. \(GM = 39.478 \times (4.217\times10^8)^3 / 152840^2 = 39.478 \times 7.499\times10^{25} / 2.336\times10^{10} = 1.267 \times 10^{17}\) m³/s²; \(M = 1.90 \times 10^{27}\) kg.
  2. \(GM = 39.478 \times (2\times10^4)^3 / 36000^2 = 3.158\times10^{14}/1.296\times10^9 = 2.44 \times 10^5\) m³/s²; \(M = 3.7 \times 10^{15}\) kg.
  3. The formula gives \(G(M+m)\). The Moon is 1.2% of the Earth's mass, which matters; Phobos is about \(10^{-8}\) of Mars's, which does not.
  4. Orbits measure \(GM\) directly and very precisely; \(G\) is known far less precisely, so dividing by it adds error.

Common mistake

Using the period in hours with \(G\) in SI units. Convert \(T\) to seconds and \(r\) to metres before using the formula.

Where it leads