nound a Stara verocity 4x1ou can coneweto the Stardaseto the Star as Paseto the Staras Pgaway from the Sta

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nound a Star
a verocity 4x
1ou can conewe
to the Stardas
eto the Star as Pa
seto the Staras Pg
away from the Sta

Expert Solution

Step 1

Newton's Law of Gravitation

According to Newton's Law of gravitation, the gravitational force between two objects of masses $m_{1}$ and $m_{2}$ with the distance between them $r$ is given by

$F = \frac{G m_{1} m_{2}}{r^{2}}$

Let us consider that the orbit of mass $m_{2}$ around $m_{1}$ is a circle. To remain in the orbit, the gravitational force must balance the centripetal force. If $v$ is the velocity of mass $m_{2}$ then

$\frac{m_{2} v^{2}}{r} = \frac{G m_{1} m_{2}}{r^{2}}$

Thus the velocity of the mass $m_{2}$ around $m_{1}$ is given by

$v = \sqrt{\frac{G m_{1}}{r}}$

Thus we see that the velocity of the body is independent of its mass and depends only on the distance between them.

Step 2

Let the mass around which the planets $P_{1}$ and $P_{2}$ rotate be $M$ . Let their velocities be $v_{1}$ and $v_{2}$ and the distance between them and the star be $R_{1}$ and $R_{2}$ respectively. Therefore we have

$v_{1} = \sqrt{\frac{G M}{R_{1}}} v_{2} = \sqrt{\frac{G M}{R_{2}}}$

In the question, it is given that the velocity of the planet $P_{1}$ is four times the velocity of the planet $P_{2}$ . Thus

$v_{1} = 4 v_{2}$

Now taking ratio

$\frac{v_{1}}{v_{2}} = \frac{\sqrt{\frac{G M}{R_{1}}}}{\sqrt{\frac{G M}{R_{2}}}} = \sqrt{\frac{R_{2}}{R_{1}}} \Rightarrow \frac{4 v_{2}}{v_{2}} = \sqrt{\frac{R_{2}}{R_{1}}}$

This gives

$\frac{R_{2}}{R_{1}} = {(4)}^{2} \Rightarrow R_{2} = 16 R_{1} \Rightarrow R_{1} = \frac{1}{16} R_{2}$

This says that $P_{1}$ is $\frac{1}{16}$ times closer to the star than $P_{2}$ .

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nound a Star a verocity 4x 1ou can conewe to the Stardas eto the Star as Pa seto the Staras Pg away from the Sta