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### Course: AP®︎/College Physics 1 > Unit 3

Lesson 7: Applications of circular motion and gravitation# Impact of mass on orbital speed

Example question exploring how mass impacts orbital speed.

## Want to join the conversation?

- This initially made sense to me, but then I started to wonder about what if the mass of the satellite was extremely large? For example, let's say I want to put a satellite into the same orbit that is the same mass as the sun? Does the mass still truly not matter? If it were possible to put something the mass of the sun into the same orbit as the first satellite would it not cause the earth to fall straight into that satellite? Therefore, meaning that the mass of the second satellite does matter?(15 votes)
- You make a good point, but that equation is that of an object with mass m that is orbiting another object of mass M. In the scenario you described, you are completely correct to say that if the object with mass m had more mass than the object with mass M it would no longer be orbiting the other object. In the case that you described, the object that had the mass of the sun would have mass M, because it would no longer be orbiting the Earth, the Earth would have mass m and would be orbiting it. I apologize for the 7 month-long wait to get your question answered.

Hope this helps!

-Hyrum(4 votes)

- During the derivation of the formula, at the point where Sal equates the centripetal force, gravity, with (m)(a), why is m the mass of the smaller object? How do we know it's not referring to the mass of earth?(3 votes)
- I guess the capital M is to show the mass of the object that is getting orbitted around (e.g earth), while the lowercase m is the mass of the object that is in the orbit (e.g satellite)(1 vote)

- If there is gravity in earth how does the earth's core stay in place(3 votes)
- I think it is because earth is a sphere and gravity is coming from all directions and keeping the core in place.(0 votes)

- Why did it all have to be proven to what v0 equals to find that the mass has no effect on the velocity? Wouldn't the centripetal acceleration be the same for both of the satellites, and we know that the radius is the same, so using ac = v^2/r, wouldn't v have to be the same for both satellites?(1 vote)
- I think you know now :)(1 vote)

## Video transcript

- [Instructor] A satellite of
mass lowercase m orbits Earth at radius capital R and speed
v naught as shown below. So this has mass lowercase m. An aerospace engineer decides
to launch a second satellite that is double the mass
into the same orbit. So the same orbit, so this radius is still gonna be capital R. And so this satellite,
the second satellite, has a mass of two m. The mass of Earth is M. So this is Earth right here, capital M. What is the speed, lowercase
v, of the heavier satellite in terms of v naught? And speed you can view as
the magnitude of velocity, and so that's why it's lowercase v without a vector symbol on it. And so what we're trying to figure out, the magnitude of its velocity,
in order to stay in orbit. What is lowercase v going to be equal to? So pause this video and see
if you can figure it out on your own. All right so to tackle this,
remember the whole reason why something stays in
orbit instead of just going in a straight line through
space is because there is going to be a constant magnitude
centripetal acceleration towards the center of
Earth that keeps turning I guess you could say the
satellite in this circular path. And we've seen from other
videos that the magnitude of our centripetal acceleration
is going to be equal to the magnitude of our velocity, and I'll just use this first satellite, so the magnitude of its
velocity squared divided by our radius, which in
this case is a capital R. But what determines our
centripetal acceleration? Well we can explore Newton's
law of gravitation there. So if we think about the
magnitude of the force of gravity, well that's going to be equal to G, which is the universal
gravitational constant, times the product of the two masses that have the force between them, so the product of the
mass of Earth, capital M, and the mass of this satellite, I'll just focus on this satellite for now, divided by the distance
between their center of masses squared. In this case that is
R, capital R, squared. And if you wanted the
centripetal acceleration, you would just divide
force divided by mass. Remember from Newton's second law, we know that F is equal to ma. And so if we're talking about
centripetal acceleration, it's the force of gravity
that is causing it. And so if you wanna solve
for centripetal acceleration, you just divide both sides of these by m. And so our centripetal acceleration here, if you divide our force
of gravity by lowercase m, by the mass of the satellite, our centripetal
acceleration is going to be the universal gravitational
constant times the mass of Earth divided by the radius squared. And so we could then
take this and substitute it back over here and solve for the magnitude of our velocity. So what you're going to
have is the universal gravitational constant
times the mass of Earth divided by the radius squared is equal to the magnitude of our
velocity or the speed squared divided by capital R. Now you can multiply both sides by R, and I'll swap sides as well, and you're gonna get v
naught squared is going to be equal to capital G times capital M over R. Or if you take the square
root of both sides, you get v naught is
equal to the square root of the universal gravitational constant times the mass of Earth
divided by the distance between the center of masses. Now what's interesting here is, we see this speed we need in
order to maintain this orbit in no way is it a function of
the mass of this satellite. I don't see a lowercase m
anywhere in this expression on the right-hand side. And since this is independent of the mass of the thing that is in orbit, if you double the mass,
if you go from lowercase m to two times lowercase m, it does not change the
needed orbital speed. And so what is the speed
v of the heavier satellite in terms of v naught? It's gonna be the same thing. We could write v is going
to be equal to v naught. Doesn't matter what you
do to the mass here, you're going to need
the same orbital speed.