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Question: In the figure shown below, two satellites, A and B both of mass \(m=125kg\), are moving in the simil...

In the figure shown below, two satellites, A and B both of mass m=125kgm=125kg, are moving in the similar circular orbit of radius r=7.87×106mr=7.87\times {{10}^{6}}m around the Earth but in opposite directions of rotation and hence on a collision course.
(a) Calculate the total mechanical energy EA+EB{{E}_{A}}+{{E}_{B}} of the two satellites ++ Earth system before the collision.

Explanation

Solution

Find the total energy of the two satellites. It will be twice the energy of the satellite. If the speed is zero, then the kinetic energy will be zero at that instant. Kinetic energy is defined as the energy which is associated with the motion of a body. Energy is expressed in joules. This all will help you in answering this question.

Complete answer:
Here we can see that the energy of each satellite is given as,
E=GMEm2rE=\dfrac{-G{{M}_{E}}m}{2r}
Therefore, the total energy of the two satellites will be the twice can be written as,
ET=EA+EB=2GMEm2r ET=GMEmr \begin{aligned} & {{E}_{T}}={{E}_{A}}+{{E}_{B}}=\dfrac{-2G{{M}_{E}}m}{2r} \\\ & \Rightarrow {{E}_{T}}=\dfrac{-G{{M}_{E}}m}{r} \\\ \end{aligned}
Where GG be the gravitational constant, ME{{M}_{E}} be the mass of the earth, mm be the mass of the satellite and rr be the radius of circular orbit given as,
m=125kgm=125kg
r=7.87×106mr=7.87\times {{10}^{6}}m
Substituting the values in the equation will give,
ET=(6.67×1011)×(5.98×1024)×1257.87×106=6.33×109J{{E}_{T}}=\dfrac{-\left( 6.67\times {{10}^{-11}} \right)\times \left( 5.98\times {{10}^{24}} \right)\times 125}{7.87\times {{10}^{6}}}=-6.33\times {{10}^{9}}J
We can see that the speed of wreckage will be zero immediately after the collision. Therefore it will be having no kinetic energy at the instant. Let us replace mm with 2m2m in the potential energy of the wreckage at that moment will be,
E=GME2m2rE=\dfrac{-G{{M}_{E}}2m}{2r}
Substituting the values in it will give,
ET=(6.67×1011)×(5.98×1024)×2×1252×7.87×106=6.33×109J{{E}_{T}}=\dfrac{-\left( 6.67\times {{10}^{-11}} \right)\times \left( 5.98\times {{10}^{24}} \right)\times 2\times 125}{2\times 7.87\times {{10}^{6}}}=-6.33\times {{10}^{9}}J
An object with a zero speed at a distance from the earth will simply fall towards the earth with a trajectory being towards the centre of the planet. The answer for the question has been obtained.

Note:
Gravitational energy otherwise called as gravitational potential energy is the potential energy of a massive body which is in relation to another massive body due to gravity. It is the potential energy in relation to the gravitational field, which is released when the body falls towards each other.