Confusion about Centripetal Force

[dmission]

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Hi. I saw this question in another thread here, and it totally stumped me as to why. I know you set centripetal force equal to the graviational force, but I'm not sure why and what that tells you.

Which of the following would be true concerning a spacecraft and the moon when they are in the same orbit? (Assume that netiher is using a propulsion system to maintain its orbit)
A. They both must be at the same speed (answer)
B. They bost must have the same mass
C. They both must have the same mass and speed
D. They must have different masses
​

Would appreciate any help, thanks.

 

[MD Odyssey]

Hi. I saw this question in another thread here, and it totally stumped me as to why. I know you set centripetal force equal to the graviational force, but I'm not sure why and what that tells you.

Which of the following would be true concerning a spacecraft and the moon when they are in the same orbit? (Assume that netiher is using a propulsion system to maintain its orbit)
A. They both must be at the same speed (answer)
B. They bost must have the same mass
C. They both must have the same mass and speed
D. They must have different masses
​

Would appreciate any help, thanks.

Consider the case of a moon or satellite orbiting the Earth at a particular distance. If you set the centripetal force equal to the gravitational force, you'll see that the mass of your satellite cancels out of the equation. This happens for the same reason that the gravitational force on a falling body is independent of its mass as well.

If you were asking why you set centripetal acceleration equal to the gravitational force, the simplest answer is to just look at the behavior of the system. You can see that the orbiting satellite is moving in a circle, therefore its acceleration is constantly towards the inside of the orbit, hence the acceleration must be centripetal.

Now, if you were moving in an ellipse, then things would be a lot more complicated and you would have additional terms to account for the additional change in your velocity.

 

[karafox]

The reason you set F_cp = F_g:

By definition F_cp is the net force that points radially toward the center of the orbit. There is only one force, F_g, assuming no propulsion (and circular orbit), as stated in the question.

Therefore,
F_net = F_g = F_cp.

 

[whiteshadodw]

The reason you set F_cp = F_g:

By definition F_cp is the net force that points radially toward the center of the orbit. There is only one force, F_g, assuming no propulsion (and circular orbit), as stated in the question.

Therefore,
F_net = F_g = F_cp.

yep and GMm/r^2 = m*v^2/r

v=(GM/r)^(1/2) showing that for a given orbital radius there is a specific speed regardless of the mass of the object. thus, they both have the same velocity/speed if they have the same orbit.

 

[dmission]

yep and GMm/r^2 = m*v^2/r

v=(GM/r)^(1/2) showing that for a given orbital radius there is a specific speed regardless of the mass of the object. thus, they both have the same velocity/speed if they have the same orbit.

I don't get it though, how does setting those equal show that?

 

[whiteshadodw]

I don't get it though, how does setting those equal show that?

when you set those equal and you solve for v you will see that v is proportional to k/(r^1/2)

where k = (GM)^1/2

you can verify that i get this v=(GM/r)^(1/2) from setting centripetal force to gravitational force. but G is the gravitational force constant, and M is the mass of the Earth or the planet that is being orbited. the only things that can vary are velocity and r. as you move away from a planet, the orbital velocity decreases. conversely, if you have a fast orbital velocity, then you must have a certain orbital radius. if two objects have the same orbital radius, then plug the r into v=(GM/r)^(1/2). you'll see that the velocity has to be the same. not only that, but because the velocity is the same, then speed must also be the same for the orbit, and thus frequency and period of the orbit are the same as well.

 

[plzNOCarribbean]

when you set those equal and you solve for v you will see that v is proportional to k/(r^1/2)

where k = (GM)^1/2

you can verify that i get this v=(GM/r)^(1/2) from setting centripetal force to gravitational force. but G is the gravitational force constant, and M is the mass of the Earth or the planet that is being orbited. the only things that can vary are velocity and r. as you move away from a planet, the orbital velocity decreases. conversely, if you have a fast orbital velocity, then you must have a certain orbital radius. if two objects have the same orbital radius, then plug the r into v=(GM/r)^(1/2). you'll see that the velocity has to be the same. not only that, but because the velocity is the same, then speed must also be the same for the orbit, and thus frequency and period of the orbit are the same as well.

perfect explanation. OP, when you set Fg=Fc, the small m, the mass of your orbiting object CANCEL's out. When you solve for velocity, you see that the orbital velocity, v=square root GM/r, which shows that the orbital velocity is inversely proportional to the square root of the orbital radius. Therefore, if we increase r, orbital velocity decreases, and if we decrease r, orbital velocity increases.