# Calculating mass of an orbiting body with force and acceleration

I'm new to physics, and it's a lot to take in- but there is a problem that I really can't seem to wrap my head around- finding the mass of an orbiting body, like an asteroid. I've looked around a lot and it seems to be impossible to find the mass of an object without going there and orbiting it, but why? If you can know the mass of the object that you're orbiting (for example the sun), can't you use that to discern the mass of the orbiting object (an asteroid)?

Then I thought of the formula F = ma, whereby you can find force with mass and acceleration. This is confusing to me, couldn't you rearrange it as M = F/a, and then find force and acceleration?

So really this is a two-pronged question: Can you find the mass of an object based on that of the one it is orbiting, or could you find it with force and acceleration?

I'm probably missing something big aren't I...

Edit: Thank you for the responses so far! I found a formula recently that purports to be able to find mass with only radius and velocity;

$M = L/rv$

where r= radius v= velocity M= mass L= angular momentum

To find angular momentum $L = Iw$

where I= moment of inertia (v/r) w= angular velocity (rv)/r^2

So it's a little complicated, but does it even work? I tried it on the mass of Venus, and I got it very, very wrong.

• So, you are on the right track. As long as the orbiting body is very light compared to the primary (which your examples are), you can't find it's mass from the large scale orbit. You have to see how it interacts with other small objects. – dmckee Oct 28 '15 at 1:11
• To add to dmckee's comment, look up "reduced mass" (en.wikipedia.org/wiki/Reduced_mass). You can then easily calculate by how much a small orbiting body will change the dynamics, which, for small bodies orbiting planets is extremely small. – CuriousOne Oct 28 '15 at 1:17
• Sometimes, a small body's mass can be estimated (not that accurately) using the observed size of the object and its reflectivity of sunlight. The idea here being that if it is a body made of mostly ice, it will have a high reflectivity, and you can use the density of ice as a starting point for the estimation. For an iron asteroid, the reflectivity is considerably lower. This is just for estimation purposes though. – tmwilson26 Oct 28 '15 at 13:35

Consider some small object orbiting the Earth. By small I mean that the mass of the object is so much smaller than the mass of the Earth that we can take the Earth to be fixed i.e. the object can't move the Earth by any measurable amount.

If the mass of the object is $m$, the mass of the Earth is $M$ and the distance to the object is $r$ then the gravitational force on the object is:

$$F = \frac{GMm}{r^2} \tag{1}$$

where $G$ is a constant called the gravitational constant. Now, you mention the equation for Newton's second law $F = ms$, and we can rearrange this to calculate the acceleration of our object:

$$a = \frac{F}{m} \tag{2}$$

If we take the force we calculated in equation (1) and substitute it into equation (2) we get:

$$a = \frac{\frac{GMm}{r^2}}{m} = \frac{GM}{r^2} \tag{3}$$

The mass of our object $m$ has factored out of the equation for the acceleration $a$, and that means the acceleration does not depend on the mass of the object. This is just Galileo's observation that objects with different masses fall at the same rate.

Anyhow, when we measure an orbit we are measuring the acceleration of the object. Since the acceleration doesn't depend on the mass that means the orbit doesn't depend on the mass. Therefore we can't determine the mass of the object from its orbit.

Both dmckee and CuriousOne have mentioned in comments that you can determine the orbit if the mass of the object is large enough to be comparable to the earth. That's because our equation (3) is actually only an approximation. It should be:

$$a = \frac{GM}{r^2}\frac{m}{\mu} \tag{4}$$

where $\mu$ is the reduced mass:

$$\mu = \frac{Mm}{M + m}$$

When $m \ll M$ the reduced mass is equal to $m$ within experimental error, and this gives us equation (3) so we can't measure $m$. If $m$ is comparable to $M$ the reduced mass is measurably different from $m$ and we can solve the resulting (rather complicated) equation to determine $m$.