Question

For a given density of the planet, the orbital period of a satellite near the surface of the planet of radius R is proportional to:
${\text{A}}{\text{. }}{R^{\dfrac{1}{2}}}$
${\text{B}}{\text{. }}{R^{\dfrac{3}{2}}}$
${\text{C}}{\text{. }}{R^{ - \dfrac{1}{2}}}$
${\text{D}}{\text{. }}{R^0}$

Answer 1

-Conclude the relationships between the velocity as well as the period of the satellite with the radius of the planet.
-Neglect the distance of the satellite from the planet since the given satellite is near the surface of the planet.
-Use the density of the planet in terms of the radius and gravitational acceleration and find the formula of the period in terms of the density.

Formula used:
The velocity of the satellite, $v = R\sqrt {\dfrac{g}{{R + h}}}$
Where $R$= the radius of the planet,
$g$= gravitational acceleration,
$h$ = the distance between the satellite and the planet.
By neglecting,$h$ i.e $h = 0$ we get,
$v = \sqrt {gR}$
The period of the satellite, $T = 2\pi \sqrt {\dfrac{R}{g}}$
The density of the planet, $\rho = \dfrac{{3g}}{{4\pi RG}}$, where $G$ = Gravitational constant.

Complete step by step answer:
The satellite of a planet moves around the planet in an elliptical or circular path with a velocity called the orbital velocity of the satellite and a period of revolution.
The formula of the orbital velocity of the satellite is, $v = R\sqrt {\dfrac{g}{{R + h}}} \ldots \ldots \ldots \ldots \ldots \ldots .\left( 1 \right)$
And the formula of the period of revolution of the satellite is, $T = \dfrac{{2\pi }}{R}\sqrt {\dfrac{{{R^3}{{(1 + \dfrac{h}{R})}^3}}}{g}} \ldots \ldots \ldots \ldots \ldots .\left( 2 \right)$
Where $R$ = the radius of the planet,
$g$ = gravitational acceleration,
$h$ = the distance between the satellite and the planet.
Since the satellite is near to the surface of the planet, the distance between the satellite and the planet can be neglected I.e $h = 0$.
Hence from Eq. (1) $\left( 1 \right)$ and (2) $\left( 2 \right)$, we get,
$v = \sqrt {gR} \ldots \ldots \ldots \ldots ..\left( 3 \right)$
And, $T = 2\pi \sqrt {\dfrac{R}{g}} \ldots \ldots \ldots \ldots \ldots .\left( 4 \right)$
The formula of the density of a planet $\rho = \dfrac{{3g}}{{4\pi RG}}$, [$G$ = Gravitational constant.]
$\Rightarrow \rho \times 4\pi RG = 3g$
$\Rightarrow \dfrac{R}{g} = \dfrac{3}{{4\pi \rho G}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left( 5 \right)$
Now put the value of $\dfrac{R}{g}$ in the Eq. $\left( 4 \right)$ we get,
$T = 2\pi \sqrt {\dfrac{3}{{4\pi \rho G}}}$
$\Rightarrow T\propto \dfrac{1}{{\sqrt \rho }}$
So we can see that for the nearer satellite of a planet, the period depends only on the density of the planet.
$\Rightarrow T\propto \dfrac{1}{{\sqrt \rho }} \times {R^0}$ [since ${R^0} = 1$].
So for a given density, the period of the satellite is proportional to ${R^0}$, i.e $T\propto {R^0}$ for a particular density $\left( \rho \right)$
Hence the right answer is in option $\left( D \right)$ .

Note:
-If the satellite is near to the surface of the planet, then the distance $h$ tends to zero$\left( 0 \right)$
-The period of the satellite that is very close to the surface of the planet only depends upon the average density of the planet.
-The velocity is high and the period is low if the distance of the satellite from the planet is very small. So to put a satellite very close to the planet a large amount of velocity is required.
-The velocity of the satellite is less than the escape velocity. The ratio of the two velocities is $0.707$.