Question

Two artificial satellites A and B are at a distance \[{{r}_{A}}\]and ${{r}_{B}}$above the earth’s surface. If the radius of the earth is $R$, then the ratio of their speeds will be:
A.) ${{\left[ \dfrac{{{r}_{B}}+R}{{{r}_{A}}+R} \right]}^{1/2}}$
B.) ${{\left[ \dfrac{{{r}_{B}}+R}{{{r}_{A}}+R} \right]}^{2}}$
C.) ${{\left[ \dfrac{{{r}_{B}}}{{{r}_{A}}} \right]}^{2}}$
D.) ${{\left[ \dfrac{{{r}_{B}}}{{{r}_{A}}} \right]}^{1/2}}$

Answer 1

Hint: The relation between the velocity of a satellite with its distance from the surface of the earth can be found out by equating that the centripetal force for its circular motion is provided for by the gravitational force of the earth. Then this relation can be used to find out the ratio of the speeds of the two satellites.
Formula used:
Gravitational force on a satellite at a distance $r$ from the surface of the earth is given by
${{F}_{g}}=\dfrac{G{{M}_{E}}m}{{{\left( {{R}_{E}}+r \right)}^{2}}}$
where ${{F}_{g}}$ is the magnitude of gravitational force, ${{M}_{E}}$ is the mass of the earth, $m$ is the mass of the satellite, ${{R}_{E}}$ is the radius of the earth, and $G$ is the universal gravitational constant equal to $6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$.
The magnitude for centripetal force ${{F}_{C}}$ on a body of mass $m$ moving at a velocity $v$ in a circular path of radius $R$is given by
${{F}_{C}}=\dfrac{m{{v}^{2}}}{R}$

Complete step by step answer:
A satellite can be considered to be moving in a circular path around the earth. We will solve this problem by finding a relation between the velocity and distance of the satellite from the center of the earth and thus compare the ratios of this relation for the two satellites.
For finding out the relation, we will first find out the magnitude of the gravitational force on the satellite due to the earth and then equate it to the magnitude of centripetal force on the satellite. The gravitational force provides the necessary centripetal force to the satellite.
Gravitational force on a satellite at a distance $r$ from the surface of the earth is given by

${{F}_{g}}=\dfrac{G{{M}_{E}}m}{{{\left( {{R}_{E}}+r \right)}^{2}}}$--(1)

where ${{F}_{g}}$ is the magnitude of gravitational force, ${{M}_{E}}$ is the mass of the earth, $m$ is the mass of the satellite, ${{R}_{E}}$ is the radius of the earth, and $G$ is the universal gravitational constant equal to $6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$.
The magnitude for centripetal force ${{F}_{C}}$ on a body of mass $m$moving at a velocity $v$ in a circular path of radius $R$ is given by

${{F}_{C}}=\dfrac{m{{v}^{2}}}{R}$
For a satellite, the radius of the circular path will be the sum of the radius of the earth and the distance of the satellite from the center of the earth. Hence, $R={{R}_{E}}+r$
$\therefore {{F}_{C}}=\dfrac{m{{v}^{2}}}{{{R}_{E}}+r}$ --(2)

Now, we will equate the magnitudes of the centripetal force and the gravitational force.

Hence, ${{F}_{C}}={{F}_{g}}$.
Therefore, using (1) and (2), we get,

$\dfrac{m{{v}^{2}}}{{{R}_{E}}+r}=\dfrac{G{{M}_{E}}m}{{{\left( {{R}_{E}}+r \right)}^{2}}}$
$\therefore {{v}^{2}}=\dfrac{G{{M}_{E}}}{{{R}_{E}}+r}$

Now, since $G$(Universal gravitational constant) and ${{M}_{E}}$(mass of the earth) are constants.
$\therefore {{v}^{2}}\propto \dfrac{1}{{{R}_{E}}+r}$
Putting square root on both sides,

$\therefore \sqrt{{{v}^{2}}}\propto \sqrt{\dfrac{1}{{{R}_{E}}+r}}$
$\therefore v\propto {{\left( {{R}_{E}}+r \right)}^{-\dfrac{1}{2}}}$ --(3)
Now, let us analyze the question.
Let the two satellites be A and B respectively.
The radius of the earth is $R$.
Distances of satellites A and B from the surface of the earth are ${{r}_{A}}$and ${{r}_{B}}$respectively.
Velocities of the satellites A and B be ${{v}_{A}}$and ${{v}_{B}}$respectively.
Therefore plugging this information in (3), we get,
$\dfrac{{{v}_{A}}}{{{v}_{B}}}=\dfrac{{{\left( R+{{r}_{A}} \right)}^{-\dfrac{1}{2}}}}{{{\left( R+{{r}_{B}} \right)}^{-\dfrac{1}{2}}}}$
$\therefore \dfrac{{{v}_{A}}}{{{v}_{B}}}=\dfrac{{{\left( R+{{r}_{B}} \right)}^{\dfrac{1}{2}}}}{{{\left( R+{{r}_{A}} \right)}^{\dfrac{1}{2}}}}$
$\therefore \dfrac{{{v}_{A}}}{{{v}_{B}}}={{\left[ \dfrac{{{r}_{B}}+R}{{{r}_{A}}+R} \right]}^{\dfrac{1}{2}}}$
Hence, the required ratio of the speeds will be${{\left[ \dfrac{{{r}_{B}}+R}{{{r}_{A}}+R} \right]}^{1/2}}$.
Hence, the correct option is A) ${{\left[ \dfrac{{{r}_{B}}+R}{{{r}_{A}}+R} \right]}^{1/2}}$.

Note: Satellites form a very important part on the topic of gravitation. A student must properly know the relations and their derivations for velocity, time period and energy. This is imperative for the point of view of all examinations.
A very common mistake that could happen here is that the student might think that since the paths of orbits are elliptical, so centripetal force with a constant value of radius cannot be applied. However, practically, the elliptical orbits are considered to be circular in nature for ease of calculating the various physical variables of the satellite.
In questions like these where the final answer is in the form of a ratio, it is best to find out a general relation in terms of the required physical variables and not go in for calculating the exact values by plugging in all values of constants and other variables since at the end, they will anyway cancel out as the answer is in the form of a ratio. This saves a lot of time and effort on the student’s part.