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Two satellites are orbiting around the earth in circular orbits of the same radius. One of them is 10 times greater in mass than the other. Their period of revolutions is in the ratio:
(A) $ \text{1}00:\text{1} $
(B) $ \text{1}:\text{1}00 $
(C) $ \text{1}0:\text{1} $
(D) $ \text{1}:\text{1} $

Last updated date: 22nd Feb 2024
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IVSAT 2024
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Hint: The formula of time period of satellite around earth is
 $ \text{T}=2\pi \sqrt{\dfrac{{{\left( \text{R}+\text{x} \right)}^{3}}}{\text{GM}}} $
The satellites are revolving in orbits of the same radius, so x is the same for both. Also, T depends on the mass of earth but not on the mass of satellites. So, by comparing T for both satellites, the ratio can be calculated.

Complete step by step solution
We know that time period of satellite around the earth is given by:
 $ \text{T}=2\pi \sqrt{\dfrac{{{\left( \text{R}+\text{x} \right)}^{3}}}{\text{GM}}} $
Where R= radius of earth
x=distance of satellite from surface of earth
G=Gravitational constant
M=Mass of earth.
Here, we can see that time period depends upon the mass of the earth and is independent of the mass of satellites.
As the radius is also same for both the satellites, therefore the ratio of time periods of two satellites is given by:
 $ \dfrac{{{\text{T}}_{1}}}{{{\text{T}}_{2}}}=\dfrac{1}{1} $
 $ {{\text{T}}_{1}}:{{\text{T}}_{2}}=1:1 $
Correct option is (D).

The satellite is revolving around the earth. The gravitational force of the earth provides the required centripetal force.
 $ \dfrac{\text{m}{{\text{v}}^{2}}}{\text{r}}=\dfrac{\text{GMm}}{{{\text{r}}^{2}}} $
Here m=mass of satellite
M=mass of earth
So, the mass of the satellite gets cancelled.
We get
 $ \begin{align}
  & {{\text{v}}^{2}}=\dfrac{\text{GMm}}{{{\text{r}}^{2}}}\times \dfrac{\text{r}}{\text{m}} \\
 & {{\text{v}}^{2}}=\dfrac{\text{GM}}{\text{r}} \\
 & \text{v=}\sqrt{\dfrac{\text{GM}}{\text{r}}} \\
\end{align} $
Now, Time period $ \text{T}=\dfrac{2\pi \text{r}}{\text{v}} $
 $ =\dfrac{2\pi \text{r}}{\sqrt{\dfrac{\text{GM}}{\text{r}}}} $
 $ \text{T}=2\pi \sqrt{\dfrac{{{\text{r}}^{3}}}{\text{GM}}} $
In our question, we have taken
 $ \text{r=R}+\text{x} $
Where r = distance of satellite from the center of the earth
R=radius of earth
x=distance of satellite from surface of the earth.

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