Question

A sky lab of mass $m{\text{ kg}}$ is first launched from the surface of the earth in a circular orbit of radius $2R$ (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius $3R.$ The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are
(A) $\dfrac{3}{4}mgR,\dfrac{{mgR}}{6}$
(B) $\dfrac{3}{4}mgR,\dfrac{{mgR}}{{12}}$
(C) $mgR,mgR$
(D) $2mgR,mgR$

Answer 1

Hint We are provided with the radius of the two orbits and the mass of the skylab. We have to find the minimum energy required to place the lab in the first orbit and energy required to shift the lab from first orbit to the second orbit. Find the energy required to place the satellites into the given orbits and then find the difference in energy required to place the two satellites into the given orbits that will be the minimum energy required to shift the lab from first orbit to the second orbit.

Complete step by step answer
Artificial satellites are objects that were invented by humans, we set them into orbits around the earth which revolve around the earth. The motion of those satellites around the earth is very similar to the motion of planets around the Sun and hence Kepler’s laws of planetary motion are applicable to the artificial satellites also. Moon is the only natural satellite of the earth. We launch artificial satellites for practical use in fields like telecommunication, geophysics and meteorology.
The total energy of satellite is given by the formula:
$E = - \dfrac{{GMm}}{{2R}}$
Where,
E is the energy of the satellite
G is the gravitational constant
M is the mass of earth
R is the radius of the earth.
m is the mass of the satellite
The total energy of the satellite is the sum of kinetic and potential energy of the satellite
Given that,
Mass of the sky lab is $m{\text{ kg}}$
At first the skylab is launched in the orbit of radius $2R$
Then the sky lab was shifted to the orbit of radius $3R.$
We have to find,
The minimum energy required to place the lab in the first orbit =?
The minimum energy required to shift the lab from first orbit to the second orbit =?
From above discussion,
The total energy of satellite at is:
$\Rightarrow T.E = - \dfrac{{GMm}}{{2R}}{\text{ }} \to {\text{1}}$
Actually the law of gravitation derived many equations regarding the gravitation, but we know only a few.
One of the equation of laws of gravitation is
$\Rightarrow g = \dfrac{{GM}}{{{R^2}}}$
$\Rightarrow g{R^2} = GM{\text{ }} \to 2$
Where,
G is the gravitational constant
M is the mass of earth
R is the radius of the earth.
The total energy required to place the lab in the first orbit of radius $2R$
$\Rightarrow T.{E_1} = - \dfrac{{GMm}}{{2(2R)}}{\text{ }} \to {\text{3}}$
Substitute equation 2 in 3 we get,
$\Rightarrow T.{E_1} = - \dfrac{{gRm}}{4}$
The minimum energy required to place the lab in the first orbit is derived by subtracting the potential energy of the satellite at the surface of the earth from the total energy required to place the lab in the first orbit.
The potential energy of the satellite at earth surface is
$\Rightarrow P.E = - \dfrac{{GMm}}{R}$
Then,
$\Rightarrow {E_1} = - \dfrac{{gRm}}{4} - \left( { - \dfrac{{GMm}}{R}} \right)$
Substitute equation 2 in above equation,
$\Rightarrow {E_1} = - \dfrac{{gRm}}{4} - ( - gRm)$
Take LCM
$\Rightarrow {E_1} = - \dfrac{{gRm}}{4} + \left( {\dfrac{{4gRm}}{4}} \right)$
$\Rightarrow {E_1} = \dfrac{{3gRm}}{4}{\text{ }} \to {\text{4}}$
The minimum energy required to place the lab in the second orbit of radius $3R$
$\Rightarrow {E_1} = - \dfrac{{GMm}}{{2(3R)}}{\text{ }} \to 5$
Substitute equation 2 in 5 we get,
$\Rightarrow {E_1} = - \dfrac{{gRm}}{6}{\text{ }} \to 6$
The minimum energy needed to shift the lab from first orbit to the second orbit is given by the difference of energy required to place the lab in the second orbit from energy required to place the lab in the second orbit
$\Rightarrow {E_S} = - \dfrac{{gRm}}{6} - \left( { - \dfrac{{gRm}}{4}} \right)$
$\Rightarrow {E_S} = - \dfrac{{gRm}}{6} + \dfrac{{gRm}}{4}$
Take LCM to add both the terms
$\Rightarrow {E_S} = - \dfrac{{2gRm}}{{12}} + \dfrac{{3gRm}}{{12}}$
$\Rightarrow {E_S} = \dfrac{{gRm}}{{12}}$
From above solution,
The minimum energy required to place the lab in the first orbit is $\dfrac{{3gRm}}{4}$
The minimum energy required to shift the lab from first orbit to the second orbit is $\dfrac{{gRm}}{{12}}$

Hence the correct answer is option (B) $\dfrac{3}{4}mgR,\dfrac{{mgR}}{{12}}$

Note To understand this problem students must know the concepts such as Kepler’s law, how satellites are launched and the derivation of formula of energy of an orbiting satellite.