AP®︎/College Physics 1
- Introduction to gravity
- Gravity for astronauts in orbit
- Would a brick or feather fall faster?
- Acceleration due to gravity at the space station
- Space station speed in orbit
- Gravitational field strength
- Comparing gravitational and inertial mass
- Impact of mass on orbital speed
- Gravity and orbits
- Newton's law of gravitation review
Example question exploring how mass impacts orbital speed.
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- 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!
- 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)
- [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.