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How do we weigh a planet?

My friend asked me how do scientists weigh a planet. He is not from science background so I need to make him understand with simple analogy. How can I explain to him?

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The other answer posted is correct, but let me try to explain it a bit more fully, and actually correct you on your terms. I did an interview on this topic for a podcast literally just two days ago, so it's fresh in my mind.


Answer: The way we determine an astronomical object's mass is through Newton's form of Kepler's Third Law, which relates the distance another object orbits the object in question to the distance it orbits from that object, and some fundamental math and physics constants. Doing this requires that there be a smaller, "test" object in orbit of the object whose mass you want to measure.

Conceptually, this means that we can only measure an object's mass via that mass's effect on another object.


What does that mean? Kepler's Third Law states that the time it takes one object to orbit another, squared, is proportional to the semi major axis of its orbit around that object, cubed. We often use that in our solar system in units of distance in A.U. (1 A.U. = Earth's average distance from the Sun) and Earth years. So, before we knew how big an A.U. is in km, we still knew that Jupiter was about 5.2 A.U., and it took 11.9 Earth years to go around the sun.

Kepler's Third Law was formed in 1619, but it took Newton's Law of Gravitation in 1687 to provide the physics behind it. That physics changed the "is proportional to" to "equals," and added 4π2/G*M into the formula. We know what 4 means, we know what π means. M is the mass of the big object that something is orbiting. It's what we want to solve for. And, assuming we can observe an orbiting object, we know its period and distance from the main object we want to solve for.

So then, all we have to do is figure out G, the gravitational constant. Gravity is really weak, so it's really hard to measure, and it took nearly 100 years for it to be calculated kinda accurately, and the first real measurement that people like (done by Cavendish) didn't happen until 1798.

With a value for "big G" as it's often called in physics class, you can now calculate any object's mass, in space. So long as it has another object orbiting it. That smaller object is required so we can use Kepler's Third Law to observe that object's orbit around the big one we want to measure. If we don't have that smaller object in orbit, we can't* measure the big one's mass. So, our first reasonable measurements of Pluto's mass only came once its main satellite, Charon, was discovered. Only asteroids with moons have reasonable mass estimates.

*There are slightly other ways to do this. One is to put an artificial satellite in orbit, so once Dawn orbited Vesta, and then Ceres, we could get their masses. Another way is a gravitational slingshot around an object, like sending a probe past Venus, we can watch how much Venus deflects the probe and get its mass. We can also watch for gravitational perturbations in a system caused by smaller masses and figure out what those smaller masses must be to cause those perturbations, but this is much harder (but led to some early estimates of the mass of Venus).

The bottom-line is still that, conceptually, you can only do this by measuring the object's (mass's) effects on another object.


Terms: Mass is a fundamental property of matter, it's a way to think of how much "stuff" is there. It has the familiar units of grams (or kilograms) or pounds. Weight is different. Weight is how mass acts in the presence of an acceleration, like gravity. It has the metric units of Newtons.

In the presence of a basically-the-same gravitational acceleration – Earth's surface – mass and weight are proportional because the gravitational acceleration is the same. So my mass – which doesn't change (much) – is going to mean I weigh the same wherever I am on Earth's surface. If my mass is 100 kg, then I will weigh about 1000 N on Earth. If I go to the Moon, my mass is still 100 kg, but I will be in a different gravitational field and I will weigh about 160 N.

Because my weight and mass are proportional practically anywhere on Earth, we use the terms "mass" and "weight" completely interchangeably, but we still use them wrong. In fact, when I did the podcast recording I mentioned above, we had to do up to seven takes of just one sentence because I kept using the terms wrong, even though I was explaining how to use them correctly!

This is important because your question is actually phrased wrong: The question is, "How do we determine the mass of a planet?"

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To weigh a planet, scientists need to know two things:

  • How long it takes objects to orbit the planet
  • How far away those objects are from the planet.

The time it takes an object to orbit a planet depends on its distance from the planet and the planet’s mass.

NASA Space Place: How Do We Weigh Planets?

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    $\begingroup$ You need to make it clearer that this only applies to planets (with satellites) in our own solar system. $\endgroup$
    – ProfRob
    Commented Sep 27, 2020 at 9:52
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    $\begingroup$ @RobJeffries Why should that not apply to exoplanets? $\endgroup$
    – jng224
    Commented Sep 27, 2020 at 10:05
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    $\begingroup$ @Jonas The mass of the star is necessary, as it's the dominant factor in Kepler's law relating the period with the semi-major axis. Once you have the semi-major axis, the velocity amplitude of the star is then also needed to estimate the planet's mass. $\endgroup$ Commented Sep 27, 2020 at 10:36
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    $\begingroup$ @Jonas We can't observe satellites going around exoplanets. $\endgroup$
    – ProfRob
    Commented Sep 27, 2020 at 15:21
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    $\begingroup$ @Jonas Zibadawa is explaining that knowing the orbital period and semi-major axis of an exoplanet does not give you it's mass. It gives the mass of the star it is orbiting. The masses of exoplanets come from measuring velocities or from transit timing variations. For massive planets in wide orbits it is possible to estimate the planet mass with a very careful fit to the projected orbit (if the distance is known). $\endgroup$
    – ProfRob
    Commented Sep 27, 2020 at 15:46

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