# An intergalactic spaceship arrives at a distant planet that rotates on its axis with a period of T. The spaceship enters a geosynchronous orbit at a distance of R.

1. Write an expression from the given data to calculate the mass of the planet concerning G and the variables given in the statement.
2. Also calculate the mass of the planet in Kg if T=26 hours and R=2.1X10^8m.

This problem aims to familiarize us with the objects revolving around a specific pivot point. The concepts required to solve this problem are mostly related to centripetal force, centripetal acceleration and orbital velocity.

According to the definition, centripetal force is the force acting on an object rotating in a circular orientation, and the object is pulled towards the axis of rotation also known as the center of curvature.

The formula for Centripetal Force is shown below:

$F = \dfrac{mv^2}{r}$

Where $m$ is the mass of the object given in $Kg$, $v$ is the tangential velocity in $m/s^2$ and $r$ is the distance of the object from the pivot point such that if the tangential velocity doubles, the centripetal force will be increased four times.

Another term to be aware of is orbital velocity, which is the velocity fine enough to induce a natural or unnatural satellite to stay in orbit. Its formula is:

$V_{orbit} = \sqrt{\dfrac{GM}{R}}$

Where $G$ is the gravitational constant,

$M$ is the mass of the body,

$R$ is the radius.

The information given in the problem statement is:

The time period of spaceship $T = 26\space hours$,

The distance of the spaceship from the axis $R = 2.1\times 10^8\space m$.

For finding the general expression for the mass of the planet, we will be using the formula of centripetal gravitational force because it provides the necessary centripetal acceleration as:

$F_c=\dfrac{GMm}{R^2}………………..(1)$

Centripetal acceleration is given as :

$a_c = \dfrac{v^2}{R}$

Also from newtons second equation of motion:

$F_c = ma_c$

$F_c = m(\dfrac{v^2}{R})$

Substituting the value of $F_c$ in equation $(1)$:

$\dfrac{GMm}{R^2} = m (\dfrac{v^2}{R})$

Simplifying the equation gives us:

$v = \sqrt{\dfrac{GM}{R}}$

Where $v$ is orbital velocity, also:

$v = \dfrac{total\space distance}{time\space taken}$

Since the total distance covered by the spaceship is circular, it will be $2\pi R$. This gives us:

$v = \dfrac{2\pi R}{T}$

$\dfrac{2\pi R}{T} = \sqrt{\dfrac{GM}{R}}$

Squaring on both sides:

$(\dfrac{2\pi R}{T})^2 = (\sqrt{\dfrac{GM}{R}})^2$

$\dfrac{4\pi^2 R^2}{T^2} = \dfrac{GM}{R}$

Rearranging it for $M$:

$M = (\dfrac{4\pi^2}{G}) \dfrac{R^3}{T^2}$

This is the general expression to find the mass of the planet.

Substituting the values in the above equation to find the mass:

$M = (\dfrac{4\pi^2}{6.67\times 10^{-11}}) \dfrac{(2.1\times 10^8)^3}{(26\times 60\times 60)^2}$

$M = (\dfrac{365.2390\times 10^{24+11-4}}{6.67\times 876096})$

$M = 6.25\times 10^{26}\space kg$

## Numerical Result

The expression is $M=(\dfrac{4\pi^2}{G}) \dfrac{R^3}{T^2}$ and the mass of the planet is $M=6.25\times 10^{26}\space kg$.

## Example

A $200 g$ ball is revolved in a circle with an angular speed of $5 rad/s$. If cord is $60 cm$ long, find $F_c$.

The equation for centripetal force is:

$F_c = ma_s$

$F_c = m \dfrac{v^2}{r} = m \omega^2 r$

Where $\omega$ is the angular velocity, substituting the values:

$F_c = 0.2\times 5^2\times 0.6$

$F_c = 3\space N$