Question

A 6480 kg satellite is in a circular earth orbit that has a radius of 2.2x10^7 m. A net force must act on the satellite to make it change to a circular orbit that has a radius of 7.55x10^6 m. What work must the net external force do?

Answer 1

\[W=-U=\int\limits_{R_{1}}^{R_{2}} F_{g}dr=GMm\int\limits_{R_{1}}^{R_{2}}\frac{dr}{r^2}=GMm\left ( \frac{1}{R_1}-\frac{1}{R_{2}} \right )\]now by substituting the values we have:
\[W=(6.67\times 10^{-11})(5.98\times 10^{24})(64820)(\frac{1}{2.2 \times 10^{7}} - \frac{1}{7.55 \times 10^{6}})=-2.25 \times 10^{12} J\]

Answer 2

Dont you have to have KE in there 1/2mv*2 to find the velocity of each and then due final minus initial?

Answer 3

Looks correct. \[F=-\frac{dU}{dR}\]Giving,\[-dU=F dR\]But F dR is just work, dW. Thus,\[-dU=dW\]We know that,\[-dU=F_{g}dr\]So just integrate to get the total work as @ParsaN has done.

Answer 4

I think @Taylorjoy123 is right. the work we calculated is due to the change of position as U is a function of r. but if we want the satellite to stay in an circular orbit we have to increase its speed as we displace it to a smaller radius. otherwise it will fall.\[W=\Delta KE-\Delta U=\frac{1}{2}m(v_{2}^2-v_{1}^2)+GMm\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )\]we know that:\[\frac{v^2}{R}=g=\frac{GM}{R^2}\rightarrow v^2=\frac{GM}{R}\]so we have:\[W=\Delta KE-\Delta U=\frac{1}{2}GMm\left ( \frac{1}{R_{2}}-\frac{1}{R_{1}} \right )+GMm\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )\]\[W=\frac{1}{2}GMm\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )\]its half of what we calculated first.

Answer 5

@ParsaN I disagree. We are imagining a force which acts to constrain the satellite as it transitions from one orbit to another. It will of course increase in speed as it goes to a smaller orbital radius but this is due to the (positive) work that gravity does on the object. Our force of constraint does no work on the satellite, it just guides it into the correct orbital shape. The change in potential energy will automatically give the satellite the correct orbital velocity. Your second analysis must be incorrect by the work-energy theorem. \[W=\Delta KE\]So, how can\[W=\Delta KE - \Delta U\] be true? This implies,\[W=W-\Delta U\]\[-\Delta U=0\]

Answer 6

I got what you say. but just half if this released energy is needed to reach the appropriate speed. where does the other half go?

Answer 7

@ParsaN, I see your point. We should have:\[\Delta KE=-\Delta U\]\[\frac{1}{2}mGM(\frac{1}{R_2}-\frac{1}{R_1})=-(-\frac{GmM}{R_2}-\frac{-GmM}{R_1})\]Giving,\[\frac{1}{2}(\frac{1}{R_2}-\frac{1}{R_1})=\frac{1}{R_2}-\frac{1}{R_1}\]I know where my confusion is stemming from. I am assuming gravity is the only force doing work on this object. This cannot be the case. It is the "net external force" which is doing work here. With this in mind, I agree with your second analysis. We cannot expect energy conservation here and my objection to your second analysis assumes work only done by gravity.

Answer 8

I think so. Im sorry I had a mistake in the second solution, its:
\[W_{ext}=\Delta K+\Delta U\]not:\[W=\Delta K-\Delta U\]