Question

An Earth satellite of mass 'm' revolves at a height 'h' from the surface of the Earth. If 'R' is the radius and 'g' is the acceleration due to gravity at the surface of the Earth, then the velocity of the satellite in its orbit is
A) $\dfrac{{g{R^2}}}{{R + h}}$
B) $\sqrt {gR}$
C) $\sqrt {\dfrac{{gh}}{{R + h}}}$
D) $\sqrt {\dfrac{{g{R^2}}}{{R + h}}}$

Answer 1

We can check the forces that are equal to each other and then calculate the required value of velocity of satellite in its orbit using those equations.
${F_{grav}} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$ expression for gravitational force where G is gravitational constant, m denotes masses of different objects exerting force on one another and r is the distance between them
Expression for centripetal force is given as $\dfrac{{m{v^2}}}{r}$, where, m is mass of the object, v, its velocity and r is its radius.
From newton’ s second law, we have, F = ma but if the body is under gravitational pull, the acceleration experienced is due to gravity and the expression becomes F = mg

Complete Step by step answer: Given:
Mass of satellite = m
Height from the surface of the earth = h
Radius of the Earth = R
Acceleration due to gravity at the surface of the Earth = g
Let the velocity of the satellite in its orbit be v.

According to newton’s law of gravitation, the gravitational pull on the acting on the satellite is given as:
Gravitational force is directly proportional to the product of two masses (exerting force on each other) and inversely proportional to the square of the distance between them.
${F_{grav}} \propto \dfrac{{{m_1}{m_2}}}{{{r^2}}}$
The gravitational constant G will be used as proportionality constant here, so:
${F_{grav}} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$
Here, ${m_1}$ is the mass of earth ‘M’ and ${m_2}$ is the mass of satellite ‘m’ and r is the distance between them given by ‘R+h’. Substituting the values we get:
${F_{grav}} = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$
The centripetal force required by the satellite to rotate is given as:
${F_C} = \dfrac{{m{v^2}}}{{R + h}}$ where,
m is the mass of the satellite, v is the velocity in its orbit and ‘R+h’ is the distance from the centre of the earth.
Now, in equilibrium, the required centripetal force to rotate is provided by this gravitational pull to the satellite.
$\Rightarrow {F_C} = {F_{grav}} \\\ \Rightarrow \dfrac{{m{v^2}}}{{R + h}} = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}} \\\$
Calculating the value of velocity of satellite (v) in its orbit:
$\dfrac{{m{v^2}}}{{R + h}} = \dfrac{{GMm}}{{{{\left( {R + h} \right)}^2}}} \\\ {v^2} = \dfrac{{GM}}{{R + h}} \\\ \Rightarrow v = \sqrt {\dfrac{{GM}}{{R + h}}} \\\$
We can find the relationship between gravitational constant G and acceleration due to gravity g:
According to newton’s law of gravity, the gravitational force on earth (bwtween earth and satellite) with radius R will be:
$F = \dfrac{{GMm}}{{{R^2}}}$
This is provided by the force acting on satellite due to the acceleration due to gravity F = mg
$\Rightarrow mg = \dfrac{{GMm}}{{{R^2}}} \\\ \Rightarrow g = \dfrac{{GM}}{{{R^2}}} \\\ \Rightarrow GM = g{R^2} \\\$
Substituting this in the value of velocity, we get:
$v = \sqrt {\dfrac{{g{R^2}}}{{R + h}}}$

Therefore, the velocity of the satellite in its orbit is $\sqrt {\dfrac{{g{R^2}}}{{R + h}}}$ and the correct option is (D).

Note: The speed of the satellite in its orbit is independent of its mass but depends on the mass and radius of earth around which it revolves.
The radius of the satellite will be measured from the centre of the earth only.Note that the gravitation is the force of attraction between the bodies of the universe whereas gravity is the earth’s gravitational pull on the body.