Question

A satellite is orbiting very close to the planet. Its time period depends only upon:
A) The density of the planet.
B) Mass of the planet.
C) The radius of the planet.
D) Mass of the satellite.

Answer 1

Hint:The formula of the time period of the satellite can be used here to get the correct answer for this problem we should as observe that in this case the satellite is revolving very close to the planet this condition will also have a huge impact on the time period of the satellite revolving around the planet.

Formula used: The formula of the time period of the satellite is given by $T=2\pi \sqrt{{\displaystyle \frac{r^3}{GM}}}$ where r is the distance between the centre of the earth and the satellite G is the gravitational constant and M is the mass of the earth.

Complete step by step answer:
The formula for the time period is given by $T=2\pi \sqrt{{\displaystyle \frac{r^3}{GM}}}$ where r is the distance between the centre of the planet to the satellite revolving around the earth, M is the mass of the earth and G is the gravitational constant.
As the density of a sphere is given by
$$\begin{gathered} \sigma ={\displaystyle \frac{M}{V}} \\ \sigma ={\displaystyle \frac{M}{\left({\displaystyle \frac{4}{3}}\pi r^3\right)}} \end{gathered}$$
As the satellite is revolving very close to the planet, therefore the radius of the planet is taken as equal to the distance between the center of the planet and the satellite.
Density in terms of mass,
$$\begin{gathered} \sigma ={\displaystyle \frac{M}{\left({\displaystyle \frac{4}{3}}\pi r^3\right)}} \\ M=\sigma \cdot \left({\displaystyle \frac{4}{3}}\pi r^3\right) \end{gathered}$$
Replacing this value of mass in the formula of the time period of the satellite we get,
$$\begin{gathered} T=2\pi \cdot \sqrt{{\displaystyle \frac{r^3}{GM}}} \\ T=2\pi \cdot \sqrt{{\displaystyle \frac{r^3}{G\left[\sigma .{\displaystyle \frac{4}{3}}\left(\pi r^3\right)\right]}}} \\ T=2\pi \cdot \sqrt{{\displaystyle \frac{3}{G.\sigma .\left(4\pi \right)}}} \\ T=\sqrt{{\displaystyle \frac{3\cdot \pi }{G.\sigma }}} \end{gathered}$$
So the time period of the satellite while revolving around the planet very closely to the planet depends upon the density of the planet

hence option C is the correct answer for this problem.

Note: Students should remember the formula for the time period of the satellite as in solving these types of the problems the formula of the time period is very important. In the formula we have taken the term r as the distance from the centre of the planet to the satellite and while calculating the density of the planet, the volume of the planet also has the radius term r. Now here we need to observe that the satellite is revolving very close to the planet therefore the radius of the planet is taken as same as the distance between the centre of planet to the satellite.