How do you calculate the altitude and velocity of a satellite in a geosynchronous orbit of mars:

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r = \sqrt[3]{\frac{GM T^{2}}{4 π^{2}}}, v = \frac{2 πr}{T}

r = \sqrt[3]{\frac{GMT}{4 π^{2}}}, v = \frac{2 πr}{T}

r = \sqrt[3]{\frac{GM T^{3}}{4 π^{2}}}, v = \frac{2 πr}{T}

r = \sqrt[3]{\frac{GM T^{2}}{π^{2}}}, v = \frac{2 πr}{T}

answer is A.

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Detailed Solution

Concept: Geosynchronous orbit: A high earth orbit that enables satellites to coincide with the earth's rotation is known as a geosynchronous orbit. This orbit is 35,786 kilometres above the equator of the planet.
The procedures below must be taken in order to determine any planet's geosynchronous orbit:
To begin with, we need to understand the centripetal force acting on the item as it is rotating. Using the formula below,

Fc = \frac{M V^{2}}{r}

The gravitational pull between two objects will then be discovered. Using the formula below,

Fc = GM \frac{m}{r^{2}}

Where r is the distance between them and M is the mass of the satellite and M is the mass of Mars.
G stands for gravity constant.
The centripetal force and gravitational force must now be equal for the satellite to remain stable in its orbit.

\frac{m v^{2}}{r} = \frac{GMm}{r^{2}}

Simplifying the equation above yields the relation shown below.

v = \frac{\sqrt{GM}}{r}

if the orbital length is 2r and the rotational time is T. Therefore, we may write the equation below.

v = \frac{2 πr}{T}

The following equation is implied;

\frac{2 πr}{T} = \frac{\sqrt{GM}}{r}

This is simplified to give us the equation that follows;

r = \sqrt[3]{\frac{GM T^{2}}{π^{2}}}

We may determine the distance using the aforementioned relation. The distance to Mars' centre is shown by this measurement. We shall deduct the radius of mass from r to determine height.
We will use the following formula to determine the velocity;

v = \frac{2 πr}{T}

We may determine the satellite's velocity in Mars' geosynchronous orbit by applying the formula above.
In light of the foregoing reasoning, it is evident how to determine a satellite's height and speed when it is in a geosynchronous orbit around Mars.
So, the correct answer is