Question

The velocity of a satellite in a parking orbit is
a) \[8Km/s\]
b) $3.1Km/s$
c) $2.35Km/s$
d) $zero$

Answer 1

Hint: Geo-stationary satellites, generally, use parking orbits as their orbit. Geo-stationary satellites have a fixed time of revolution around the earth. Kepler’s formula along with the orbital velocity formula can be used to find the answer.
Formula Used:
$T=\sqrt{\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}}$
$v=\sqrt{\dfrac{GM}{r}}$

Complete answer:
Parking orbits are used by all satellites launched as a temporary orbit. They stay in that orbit for a while and then fire their way further up into the destination orbit.
Here, we can find the velocity of satellites in the parking orbit using the simple formula of velocity of a satellite around a planet and also using Kepler’s Law.
We know that the parking orbit is, usually, the orbit of geo-stationary satellites. Also, geo-stationary orbits have a time period of $24hr$.
According to Kepler’s Law:
$T=\sqrt{\dfrac{4{{\pi }^{2}}{{r}^{3}}}{GM}}$ -----(i)
Where, $T$ is the time period in seconds, $r$ is the orbital radius, $M$ is the mass of earth.
Using this equation, we can find the radius of the parking orbit as:
$r=\sqrt[3]{\dfrac{{{T}^{2}}GM}{4{{\pi }^{2}}}}$ ----(ii)
Now, we also know that the velocity of a satellite orbiting with a velocity $v$ with an orbital radius of $r$ around a planet of mass $M$ is given by:
$v=\sqrt{\dfrac{GM}{r}}$ ----(iii)
Putting the value of $r$ from equation (ii) in equation (iii), we get:
$v=\sqrt{\dfrac{GM}{\sqrt[3]{\dfrac{{{T}^{2}}GM}{4{{\pi }^{2}}}}}}$
$\Rightarrow {{v}^{2}}=GM\times \sqrt[3]{\dfrac{4{{\pi }^{2}}}{{{T}^{2}}GM}}$
$\Rightarrow v=\sqrt[3]{\dfrac{2\pi GM}{T}}$
Now, putting the values of respective parameters, we get:
$\Rightarrow v=\sqrt[3]{\dfrac{2\pi \times 6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{24\times 3600}}$
$\Rightarrow v=3.1\times {{10}^{3}}m/s$
$\Rightarrow v=3.1Km/s$
Hence, clearly, the velocity of the satellite in the parking orbit is $3.1Km/s$.

So the answer is option (b).

Note:
Note that the $M$ mentioned in the formula is the mass of earth. It is not compulsory for the mass of the earth to be given in the question. In that case, you will have to remember the mass of earth along with the gravitational constant to solve the question.