Question

Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to ${R^{ - \dfrac{5}{2}}}$, then the square of the time period will be proportional to

A. ${T^2}$ is proportional to ${R^2}$
B. ${T^2}$ is proportional to ${R^{\dfrac{7}{2}}}$
C. ${T^2}$ is proportional to ${R^{\dfrac{3}{2}}}$
D. ${T^2}$ is proportional to ${R^{3.75}}$

Answer 1

When satellites are revolving around bigger planets those are acted upon by gravitational force due to attraction between planet and satellite. This gravitational force of attraction is the reason for the revolution as this provides the required centripetal force. By equating to centripetal force we can get velocity and then time period.
Formula used:
Time period of revolution(T) = $2\pi R/v$
${F_c} = \dfrac{{m{v^2}}}{R}$

Complete answer:
It is given that a light planet is revolving around a huge planet hence we can assume a light planet as a satellite.
Let gravitational force be ${F_g}$
It is given that
${F_g}\alpha {R^{ - \dfrac{5}{2}}}$ …eq1
This gravitational force will be responsible for centripetal force and let it be ${F_c}$
We know ${F_c} = \dfrac{{m{v^2}}}{R}$
Where m is mass of satellite, v is velocity of revolution and R is the radius of orbit.
From equation 1 we have
\eqalign{ & {F_g}\alpha {R^{ - \dfrac{5}{2}}} \cr & but{F_g} = {F_c} \cr & i.e{F_c}\alpha {R^{ - \dfrac{5}{2}}} \cr & {v^2}\alpha {R^{ - \dfrac{5}{2} + 1}} \cr & \implies v\alpha {R^{ - \dfrac{3}{4}}} \cr}
Time period of revolution(T) = $2\pi R/v$
From this we can derive relation between time period and radius of orbit
\eqalign{ & 2\pi R/v \cr & \implies T\alpha \dfrac{R}{v} \cr & \implies v\alpha {R^{\dfrac{{ - 3}}{4}}} \cr & \implies T\alpha {R^{\dfrac{7}{4}}} \cr & \implies {T^2}\alpha {R^{\dfrac{7}{2}}} \cr}

So, the correct answer is “Option B”.

Additional Information:
Here it is given as circular orbit but generally satellites revolve in elliptical orbits. Then one should replace radius with semi major axis length in every formula, let it be total energy or potential energy or time period formula. In the case of a circle , the semimajor axis will be equal to radius.

Note:
There is kepler’s third law stating the relation between the time period of rotation of a satellite and radius of the orbit. The gravitational force is proportional to inverse square of radius and we get square of time period proportional to cube of radius. One should not confuse that result with this one because here inverse square law is not followed in case of gravitational force.