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Question

A spacecraft of mass m orbits Earth at a radius R and speed $v_0$ as shown below. An aerospace engineer decides it should orbit at a radius of $\frac{2R}{3}$ instead. The mass of Earth is M.

What is the new speed, $v_n$, of the spacecraft in terms of $v_0$?

  • My answer

We can begin by saying $v_0 = ωR$ where ω is angular velocity.

Every single point on the radius of this circle has the same angular velocity. I understand that different points on this radius will have different linear velocities. Now if we were to travel 2/3 of the radius up from the centre of the circle, then at that point the linear velocity would be

$v_n$ = ω$\frac{2R}{3}$ = $\frac{2}{3}ωR$ = $\frac{2}{3}v_0$

The actual correct answer:

This mass, $m$, has a centripetal acceleration, $a_c$, which is caused by a centripetal force which in this case is the force of gravity the Earth applies to $m$. So by Netwon's second law we can say

$F_g$ = m$a_c$.

$\frac{GMm}{R^2}$ = $\frac{mv_0^2}{R}$

$v_0$ = $\sqrt\frac{GM}{R}$

$v_n$ = $\sqrt\frac{GM}{\frac{2R}{3}}$

$v_n$ = $\sqrt\frac{3}{2}$ $\sqrt\frac{GM}{R}$ = $\sqrt\frac{3}{2}v_0$

I understand the correct answer which is good but I don't understand why what I wrote was wrong. Why can't the relationship $v = ωR$ be used to find out the what the new velocity is.

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You may follow a similar derivation and find the new angular velocity,

$F_g$ = m$a_c$

$\frac{GMm}{R^2}$ = $mR\omega_0^2$

$\omega_0$ = $\sqrt\frac{GM}{R^3}$

Then, the new angular velocity is given by,

$\omega_n$ = $\sqrt\frac{GM}{(2R/3)^3}= \left(\frac{3}{2}\right)^{3/2}\omega_0$

As seen here, the new angular velocity $\omega_n\ne\omega_0$. In your approach, you assume they are the same somehow.

The new speed therefore is,

$v_n=\omega_n \frac{2R}{3} = \sqrt{\frac{3}{2}} \omega_0 R= \sqrt{\frac{3}{2}}v_0$

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  • $\begingroup$ I understand your answer but isn't is true that every point on the radius R has the same angular velocity. I learnt this from Khan Academy @time stamp: 07.43 - 09:03: khanacademy.org/science/physics/ap-physics-1/… $\endgroup$ – CubbyKushi Aug 12 at 11:19
  • $\begingroup$ That is only true for a radius rotating around center, say, a hand of a clock. But, that is not the setup in question. $\endgroup$ – Quanto Aug 12 at 11:22
  • $\begingroup$ How is the setup in the question different to a radius rotating around a centre? $\endgroup$ – CubbyKushi Aug 12 at 11:24
  • $\begingroup$ Is the difference is that, in reality there's no rigid body connecting the centre of the Earth the spacecraft of mass m? Which means when a new spacecraft is introduced in an orbit 2/3 of the radius, it's entirely possible for it to move at an entirely different angular velocity. $\endgroup$ – CubbyKushi Aug 12 at 11:33
  • $\begingroup$ That’s correct. $\endgroup$ – Quanto Aug 12 at 11:36
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The spacecraft is moving in a circular path because It's in equilibrium: The gravitational force equals to the centrifugal force. At $\frac{2}{3}R$, the gravitational force will be stronger, so we would need stronger centrifugal force as well; and it can't happen with the same angular velocity.

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