# A very simple question on motion in a circle.

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 $$v_0$$?

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

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.

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,

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

• 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/… – CubbyKushi Aug 12 at 11:19
• That is only true for a radius rotating around center, say, a hand of a clock. But, that is not the setup in question. – Quanto Aug 12 at 11:22
• How is the setup in the question different to a radius rotating around a centre? – CubbyKushi Aug 12 at 11:24
• 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. – CubbyKushi Aug 12 at 11:33
• That’s correct. – Quanto Aug 12 at 11:36

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.