Question

Two satellites, A and B, have masses m and 2m respectively. A is in circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, $\dfrac{{{T_A}}}{{{T_B}}}$ is:
A. 2
B. \[\sqrt {\dfrac{1}{2}} \]
C. 1
D. $\dfrac{1}{2}$

Answer 1

Hint: In this question, we need to determine the ratio of the kinetic energy of the two satellites such that the radius of the circular orbit of the satellite A and B are R and 2R respectively. For this, we will use the relation between the kinetic energy of the particle, mass of the particle and the radius of the circular path.

Complete step by step answer:
The mass of the satellite A is m and is traveling in a circular path of radius R.
The mass of the satellite B is 2m and is traveling in a circular path of radius 2R.
The kinetic energy of the body around the body of the mass M is given as:

$T = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{R}} } \right)^2}$ where, ‘m’ is the mass of the body which is revolving around a fixed body of mass ‘M’ in a fixed circular path of radius ‘R’ and ‘G’ is the gravitational constant.
Let the mass of the earth be “M”.
According to the question, the satellite A is of mass ‘m’ is revolving around the earth in a circular path of radius R so, the kinetic energy is given as:
$
{T_A} = \dfrac{1}{2}m{\left( {\sqrt {\dfrac{{GM}}{R}} } \right)^2} \\
  = \dfrac{{GMm}}{{2R}} - - - - (i) \\
 $
Similarly, the satellite B is of mass ‘2m’ is revolving around the earth in a circular path of radius 2R so, the kinetic energy is given as:
$
{T_B} = \dfrac{1}{2}(2m){\left( {\sqrt {\dfrac{{GM}}{{2R}}} } \right)^2} \\
 = \dfrac{{GMm}}{{2R}} - - - - (ii) \\
 $
Now, we need to evaluate the ratio of the kinetic energy of the satellites A and B so, divide the equation (i) and (ii), we get
$
\dfrac{{{T_A}}}{{{T_B}}} = \dfrac{{\left( {\dfrac{{GMm}}{{2R}}} \right)}}{{\left( {\dfrac{{GMm}}{{2R}}} \right)}} \\
   = 1 \\
 $
Hence, the ratio of the kinetic energy of the satellites A and B is 1.

So, the correct answer is “Option C”.

Note:
Half of the product of the mass of the body and the square of the velocity of the body results in the kinetic energy of the body. Here, the velocity of the body (satellite) is given as $v = \sqrt {\dfrac{{GM}}{R}} $ and the same have been used in the solution.