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Question: An asteroid of mass m is approaching earth, initially at a distance \(10{R_E}\) with speed\({v_i}\)....

An asteroid of mass m is approaching earth, initially at a distance 10RE10{R_E} with speedvi{v_i}. It hits earth with a speed vf{v_f} (RE and ME{R_{E{\text{ }}}}and{\text{ }}{{\text{M}}_E} are radius and mass of the earth), then

& A.{\text{ }}{{\text{v}}_f}^2 = v{{}_i^2} + \dfrac{{2Gm}}{{{R_E}}}\left( {1 + \dfrac{1}{{10}}} \right) \cr & B.\;{{\text{v}}_f}^2 = v{{}_i^2} + \dfrac{{2G{m_E}}}{{{R_E}}}\left( {1 + \dfrac{1}{{10}}} \right) \cr & C.\;{{\text{v}}_f}^2 = v{{}_i^2} + \dfrac{{2G{m_E}}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right) \cr & D.\;{{\text{v}}_f}^2 = v{{}_i^2} + \dfrac{{2Gm}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right) \cr} $$
Explanation

Solution

According to the law of conservation of energy, energy can neither be created nor destroyed; it only changes its form from one into another. It states that the total energy of an isolated system remains constant and the energy is conserved over the time. Use formula for the law of conservation to find the final velocity.

Complete step by step answer:
The initial energy of the asteroid is the total of the initial kinetic energy and the initial potential energy.
Therefore,
Ei=Ki+Ui{E_i} = {K_i} + {U_i}
As it is given that an asteroid of mass m is approaching earth, initially at a distance 10RE10{R_E} with speedvi{v_i}implies –
\eqalign{ & {K_i} = \dfrac{1}{2}m{v_i}^2 \cr & {U_i} = \dfrac{{GM{}_Em}}{{10{R_E}}} \cr}
Hence, Initial Energy
Ei=12mvi2−GMEm10RE .......(a){E_i} = \dfrac{1}{2}m{v_i}^2 - \dfrac{{GM{}_Em}}{{10{R_E}}}{\text{ }}.......{\text{(a)}}
(The acceleration due to gravity is always negative for an asteroid or for any free falling object)
Similarly, the final energy of the asteroid is Ef{E_f}
Ef=12mvf2−GMEmRE ........(b){E_f} = \dfrac{1}{2}m{v_f}^2 - \dfrac{{GM{}_Em}}{{{R_E}}}{\text{ }}........{\text{(b)}}
According to the law of the conversation of energy – The initial Energy is equal to the final
energy.
Ei=Ef{E_i} = {E_f}
Therefore, equate values of the equation (a) and (b)
12mvi2−GMEm10RE = 12mvf2−GMEmRE \dfrac{1}{2}m{v_i}^2 - \dfrac{{GM{}_Em}}{{10{R_E}}}{\text{ = }}\dfrac{1}{2}m{v_f}^2 - \dfrac{{GM{}_Em}}{{{R_E}}}{\text{ }}
Simplify the above equations and make vf{v_f} the subject –
12mvi2−GMEm10RE+GMEmRE = 12mvf2 \dfrac{1}{2}m{v_i}^2 - \dfrac{{GM{}_Em}}{{10{R_E}}} + \dfrac{{GM{}_Em}}{{{R_E}}}{\text{ = }}\dfrac{1}{2}m{v_f}^2{\text{ }}
Multiply both the sides of the equation with
mvi2−2GMEm10RE+2GMEmRE = mvf2 m{v_i}^2 - \dfrac{{2GM{}_Em}}{{10{R_E}}} + \dfrac{{2GM{}_Em}}{{{R_E}}}{\text{ = }}m{v_f}^2{\text{ }}
Take mass, “m” common from both the sides of the equations-
vi2−2GME10RE+2GMERE = vf2 {v_i}^2 - \dfrac{{2GM{}_E}}{{10{R_E}}} + \dfrac{{2GM{}_E}}{{{R_E}}}{\text{ = }}{v_f}^2{\text{ }}
∴vf2=vi2−2GME10RE+2GMERE \therefore {v_f}^2 = {v_i}^2 - \dfrac{{2GM{}_E}}{{10{R_E}}} + \dfrac{{2GM{}_E}}{{{R_E}}}{\text{ }}
Rearrange the above equation –
  vf2=vi2+2GmERE(1−110)\;{{\text{v}}_f}^2 = v{{}_i^2} + \dfrac{{2G{m_E}}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right)
Therefore, the required solution is – An asteroid of mass m is approaching earth, initially at a distance 10RE10{R_E} with speed vi{v_i}. It hits earth with a speed vf{v_f} (RE and ME{R_{E{\text{ }}}}and{\text{ }}{{\text{M}}_E} are radius and mass of the earth), then   vf2=vi2+2GmERE(1−110)\;{{\text{v}}_f}^2 = v{{}_i^2} + \dfrac{{2G{m_E}}}{{{R_E}}}\left( {1 - \dfrac{1}{{10}}} \right)
Hence, from the given multiple choices – the option C is the correct answer.

Note: Follow different concepts of the initial energy of the free falling objects and accordingly take the sign (positive or negative) for the acceleration due to gravity. Always apply carefully basic operations to simplify and get the required solution. Remember, the universe is itself the closed system so the total amount of energy existed in the universe remains the same.