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Question: Find the escape velocity from the earth for a 1000 kg spacecraft and find the kinetic energy it must...

Find the escape velocity from the earth for a 1000 kg spacecraft and find the kinetic energy it must have at the surface of the earth in order to escape the earth’s gravitational field. M=6×1024kgM = 6 \times {10^{24}}\,kg, R=6400kmR = 6400\,km and G=6.67×1011G = 6.67 \times {10^{ - 11}}.

Explanation

Solution

Use the formula for escape velocity of a planet to calculate the escape velocity of the spacecraft. We know that the escape velocity is independent of the mass of the body. Use the formula for kinetic energy and substitute the value of escape velocity in it.

Formula used:
ve=2GMR{v_e} = \sqrt {\dfrac{{2GM}}{R}}
Here, G is the gravitational constant, M is the mass of earth and R is the radius of earth.
K.E=12mv2K.E = \dfrac{1}{2}m{v^2}
Here, m is the mass of the body and v is the velocity.

Complete step by step answer:
We know the formula for calculating the escape velocity of the planet,
ve=2GMR{v_e} = \sqrt {\dfrac{{2GM}}{R}} …… (1)
Here, G is the gravitational constant, M is the mass of earth and R is the radius of earth.
We know that the acceleration due to gravity is can be calculated using the relation,
g=GMR2g = \dfrac{{GM}}{{{R^2}}} …… (2)
Using equation (2) in equation (1), we get,
ve=2gR{v_e} = \sqrt {2gR}
We substitute 9.8ms29.8\,m{s^{ - 2}} for g and 6400 km for R in the above equation.
ve=2(9.8m/s2)(6.4×106m){v_e} = \sqrt {2\left( {9.8\,m/{s^2}} \right)\left( {6.4 \times {{10}^6}\,m} \right)}
ve=1.254×108\Rightarrow {v_e} = \sqrt {1.254 \times {{10}^8}}
ve=11200m/sor 11.2km/s\Rightarrow {v_e} = 11200\,m/s\,\,\,{\text{or }}\,{\text{11}}{\text{.2}}\,\,km/s
Thus, the velocity of the spacecraft should be 11.2 km/s to escape from earth’s gravitational field.Now, we know the formula for kinetic energy of the body. With this initial velocity of the spacecraft, its kinetic energy will be,
K.E=12mve2K.E = \dfrac{1}{2}mv_e^2
Here, m is the mass of the spacecraft.
We substitute 1000 kg for m and 11200 m/s for ve{v_e} in the above equation.
K.E=12(1000)(11200)2K.E = \dfrac{1}{2}\left( {1000} \right){\left( {11200} \right)^2}
K.E=5.6×106J\therefore K.E = 5.6 \times {10^6}\,J

Therefore, the kinetic energy of the spacecraft is 5.6×106J5.6 \times {10^6}\,J.

Note: We could have used equation (1) to calculate the escape velocity but it involves three quantities which will be difficult for calculation. Therefore, we have substituted the value of acceleration due to gravity in the first equation. So, for the sake of calculations, students should know all the necessary formulae in gravitation.