1
$\begingroup$

Two artificial satellites are in circular earth orbits in the same plane. The orbit radius of one of the satellite is $r$ and that of the other is slightly greater than the other i.e $r + \Delta r$. What is the time interval $t$ between successive closest approach of the satellites?

My Attempt:

I found the absolute difference of their angular velocities i.e $\omega_{\text{rel}} = |\omega_1 - \omega_2|$. So, the time period of closest approach will be given by,

$$T = \frac{2\pi}{\omega_{rel}}$$

$$w_1 = \frac{\sqrt{GM}}{r^{\frac{3}{2}}}$$

$$w_2 = \frac{\sqrt{GM}}{(r + \Delta r)^{\frac{3}{2}}}$$

taking r common from the denominator,

$$\omega_2 = \frac{\sqrt{GM}}{(r^{3/2})(1 + \frac{\Delta r}{r})^{3/2}}$$

Applying the binomial approximation, I got

$$\omega_2 = \frac{\omega_1}{1 + \frac{3\Delta r}{2r}}$$

Then, taking the absolute difference and then putting in the equation, I got

$$\omega_{rel} = \omega_1\bigg(\frac{3\Delta r}{2r}\bigg)$$

$$T = 2\pi\bigg(\frac{2r}{\omega_1(3\Delta r)}\bigg)$$

But then I stopped and didn't care to substitute $\omega_1$ because the answer given has $\Delta r$ factor in the numerator.

Since the question says that the difference in radius is very small, I thought it has to do with something like differentiation and infinitesimals. Just for some experimentation, I differentiated the time period of an object performing a circular motion in the gravitational field w.r.t radius.

$$T = \frac{2\pi r^{3/2}}{\sqrt{GM}}$$

$$\frac{dT}{dr} = \frac{3\pi \sqrt r}{\sqrt GM}$$

$$dT = \bigg(\frac{3\pi \sqrt r}{\sqrt GM}\bigg)dr$$

$$\Delta T = 3\pi\Delta r\sqrt{\frac{r}{GM}}$$

And guess what, this is the correct answer! So what is going on here? What does $\Delta T$ signify here? What does it mean?

Moreover, what's wrong in my original approach?

Any hint/help would be appreciated.

$\endgroup$

closed as off-topic by John Rennie, Jon Custer, GiorgioP, Kyle Kanos, Phonon Mar 28 at 17:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Jon Custer, GiorgioP, Kyle Kanos, Phonon
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Hi Tony and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. $\endgroup$ – John Rennie Mar 25 at 8:02
  • 1
    $\begingroup$ But I have a conceptual doubt. Not just any random doubt that asks what's the problem in this line or something else. Moreover I have also shown my research and thoughts. $\endgroup$ – Tony Mar 25 at 8:11
  • $\begingroup$ Your first answer looks correct to me. You've used the fact that $\Delta r$ is small when expanded the binomial approximation to the first order. Obviously, as $\Delta r$ approaches zero, the $T$ must approach infinity: the orbits are nearly same, so it will take a long time and many revolutions for the bodies to meet. The second equation from the bottom, however, does not yield the last one (integrate the right side carefully), and when you expand the powers, discard higher-order powers of $\Delta r$, and will probably arrive at your first solution. But the first solution is much cleaner! $\endgroup$ – kkm Mar 26 at 9:50
  • $\begingroup$ I didn't get what you are trying to say. Are both the answers correct? But how can that be possible? Also, how would integrating the bottom equation lead me to the equation in my first approach? $\endgroup$ – Tony Mar 27 at 9:37