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Question: At a height \(H\) from the surface of the earth, the total energy of a satellite is equal to the pot...

At a height HH from the surface of the earth, the total energy of a satellite is equal to the potential energy of a body of equal mass at a height of 3R3R from the surface of the earth (RR=Radius of the earth). Find the value of HH.
A. RR
B. 4R3\dfrac{{4R}}{3}
C. 3R3R
D. R3\dfrac{R}{3}

Explanation

Solution

Hint: A satellite is located at a height of HH from the surface of the earth, it will have some energy. That energy is equal to the potential energy of the body of equal mass located at a height of 3R3R from the surface. By using the energy formula, the energy of the satellite and the body can be obtained and by equating those two energies the value of HH can be calculated.

Useful formula:
Total energy of satellite located in space,
T.E=−GMm2r1T.E = - \dfrac{{GMm}}{{2{r_1}}}
Where, GG is gravitational constant, MM is the mass of the earth, mm is the mass of the satellite and r1{r_1} is the distance between the centre of earth to the satellite.

Potential energy of the object,
P.E=−GMmr2P.E = - \dfrac{{GMm}}{{{r_2}}}
Where, GG is gravitational constant, MM is the mass of the earth, mm is the mass of the body and r2{r_2} is the distance between the centre of earth to the body.

Given data:
The radius of the earth is RR
The distance between the surface of the earth and the object is 3R3R
The height of satellite located from the surface of the earth is HH

Step by step solution:
Total energy of satellite located in space,
T.E=−GMm2r1T.E = - \dfrac{{GMm}}{{2{r_1}}}
Since, the distance between the centre of the earth to the satellite is r1=R+H{r_1} = R + H
Hence, T.E=−GMm2(R+H)T.E = - \dfrac{{GMm}}{{2\left( {R + H} \right)}}

Potential energy of the object,
P.E=−GMmr2P.E = - \dfrac{{GMm}}{{{r_2}}}
Since, the distance between the centre of the earth to the body is r2=R+3R{r_2} = R + 3R
Hence,
P.E=−GMm(R+3R) P.E=−GMm4R  P.E = - \dfrac{{GMm}}{{\left( {R + 3R} \right)}} \\\ P.E = - \dfrac{{GMm}}{{4R}} \\\

From the question,
T.E=P.ET.E = P.E
Substitute the values of energies in above relation, we get
\-GMm2(R+H)=−GMm4R 2(R+H)=4R R+H=2R H=2R−R H=R  \- \dfrac{{GMm}}{{2\left( {R + H} \right)}} = - \dfrac{{GMm}}{{4R}} \\\ 2\left( {R + H} \right) = 4R \\\ R + H = 2R \\\ H = 2R - R \\\ H = R \\\

Hence, the option (A) is correct.

Note: It is clear from the question, there are two objects in space. One is a satellite and the other is a body. It is given that the total energy of the satellite is equal to the potential energy of the body. By equating both the energies, the value of the height of the satellite is obtained.