Question

A satellite \[X\] of mass m orbits the Earth with a period \[T\] . What will be the orbital period of the satellite \[Y\] of mass \[2m\] occupying the same orbit as \[X\] ?
A. $\dfrac{T}{2}$
B. $T$
C. ${2^{0.5}}T$
D. $2T$

Answer 1

Hint:In order to answer the given question, the first thing we will do is that we will calculate gravitational force by equating the centripetal force and also find the velocity $\left( v \right)$ , and after that finally putting the values in the formula of time period we will find out our desired answer.

Complete step by step answer:
Let's begin by using what we already know to derive a formula for the period.
The first step is to calculate the gravitational force by equating centripetal force $\left( {\dfrac{{m{v^2}}}{r}} \right)$
The gravitational force between the satellite and the earth is equivalent to this force. Therefore
\[\]$\dfrac{{{m_X}{v^2}}}{r} = \dfrac{{G{m_X}{m_E}}}{{{r^2}}}$
From here we will equate for $v$
$v = \sqrt {\dfrac{{G{m_E}}}{r}} $
We obtain that because $t = \dfrac{d}{v}$ and the orbit around the earth is \[2\pi r\] .
The satellite's period is calculated as follows:
$P = \dfrac{{2\pi r}}{v}$
Putting the value of $v$ in the above formula;
$
  P = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{G{m_E}}}{r}} }} \\
  P = 2\pi \sqrt {\dfrac{{{r^3}}}{{G{m_E}}}} \\
 $
As you can see, the period is solely determined by the Earth's mass, orbital radius, and gravitational constant. As a result, the orbital period remains \[T\] despite the fact that the satellite's mass has no influence on it.
Hence, the correct option is: (B) $T$

Note:The orbital period (also known as the revolution period) is the time it takes for a given astronomical object to complete one orbit around another object. It is most commonly used in astronomy to describe planets or asteroids orbiting the Sun, moons orbiting planets, exoplanets orbiting other stars, and binary stars.