I have a question says that "If the orbital radius of a satellite increase four times the orbital velocity will be ..." and the answer is that the velocity will be doubled !!

We studied that orbital velocity = $\sqrt{\frac{GM}{r}} = \frac{2\pi r}{t}$ I can't figure out how the answer is that the velocity will be doubled and how in the two laws the radius is directly proportional to velocity in one of them and inversely proportional in the other!!

  • $\begingroup$ It isn’t clear what “two laws” you are referring to, but the equation you wrote is correct and tells you the right answer. $\endgroup$ – G. Smith May 27 '19 at 23:20
  • $\begingroup$ @G.Smith the equation I wrote indicates that the answer should be half not double $\endgroup$ – Mostafa Khaled May 27 '19 at 23:30
  • 1
    $\begingroup$ That’s correct. Was the question with the wrong answer in some book? Or did it come from your teacher? $\endgroup$ – G. Smith May 27 '19 at 23:54

The given answer is wrong. Multiplying $r$ by 4 divides $v$ by 2.

You have two equations for the orbital speed of a (small) body in a circular orbit. $$v=\sqrt{\frac{GM}r} \tag 1$$ and $$v=\frac{2\pi r}T \tag 2$$ where $r$ is the orbital radius, and $T$ is the orbital period.

Now from equation 1 it looks like $v$ is inversely proportional to $\sqrt r$, but it's also directly proportional to $r$ from equation 2. How can that work?

Simple! $T$ isn't independent of $r$. Let's equate the two equations. $$v^2 = \frac{GM}r = \frac{4\pi^2r^2}{T^2}$$ $$\left(\frac{GM}{4\pi^2}\right)T^2=r^3$$

You may recognise this as Kepler's 3rd law.

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enter image description here

As you can clearly see, the velocity is proportional to
$$v_{orbital} \propto \sqrt{\frac{1}{r}}$$
I'm pretty sure it should halve since it's square-root inversely proportional. Unless I'm missing something obvious or am a bit slow at this late hour (1 AM), the given answer is incorrect as it should have been "halved"

Source: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/Keporb.html

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