Question

A satellite is in a circular orbit around a planet, its period of revolution is $T$, radius of the orbit is $R$, orbital velocity $V$ and acceleration $a$ then:
A. $V = aT$ and $a = \dfrac{{{V^2}}}{R}$
B. $V = \dfrac{{2\pi R}}{T}$ and $aT$
C. $V = \dfrac{{2\pi R}}{T}$ and $a = \dfrac{{{V^2}}}{R}$
D. $V = \dfrac{1}{2}a{T^2}$

Answer 1

Recall the formula of centripetal acceleration and linear velocity. Find the relation between angular velocity and linear velocity to find the formula of linear velocity in terms of time period.

Complete step by step answer: Consider a satellite performing circular motion such that
$R$ is the radius
$T$ is the period of revolution
$V$ is linear or tangential velocity
$a$ is acceleration.
Then we know that acceleration always acts along the radius of the circular orbit. And it is called centripetal acceleration.
Centripetal acceleration is given by the formula,
$a = \dfrac{{{V^2}}}{R}$
The period of revolution is the time taken to complete one revolution.
It is given by
$T = \dfrac{{2\pi }}{\omega }$
Where,
$\omega$ is the angular velocity of the satellite
Rearranging it we can write
$\omega = \dfrac{{2\pi }}{T}$
But we also know that linear velocity is the product of angular velocity and radius of the orbit.
$\Rightarrow V = R\omega$
$\Rightarrow V = \dfrac{{2\pi R}}{T}$
Thus the centripetal acceleration is $a = \dfrac{{{V^2}}}{R}$ and linear velocity is $V = \dfrac{{2\pi R}}{T}$
Therefore, from the above explanation, the correct answer is, option (B) $V = \dfrac{{2\pi R}}{T}$ and $a = \dfrac{{{V^2}}}{R}$

Note: To solve this question, you need to know the formulas used in rotational motion and the relation between angular and linear quantities like velocity etc.
Linear acceleration always acts along the center of the circle in which the body is orbiting and is also called the centripetal or radial acceleration.
Linear velocity always acts along the tangent to the circle in which the object is moving.