Question

The mass of a spaceship is 1000kg. It is to be launched from the earth surface out into free space. The value of $g$ and $R$ (radius of earth) are $10m/{s^2}$ and $6400km$ respectively. The required energy for this work will be:
(A) $6.4 \times {10^{11}}J$
(B) $6.4 \times {10^8}J$
(C) $6.4 \times {10^9}J$
(D) $6.4 \times {10^{10}}J$

Answer 1

For solving this question, we need to evaluate the value of the escape velocity. Then using this value, we can calculate the required kinetic energy and hence the energy required for the work.

Formula used: The formulae which are used to solve this question are given by
${v_e} = \sqrt {2gR}$ , here ${v_e}$ is the escape velocity of an object to make it free of the gravitational influence of the earth, $g$ is the acceleration due to gravity, and $R$ is the radius of the earth.
$K = \dfrac{1}{2}m{v^2}$ , here $K$ is the kinetic energy of a body of mass $m$ which is moving with a speed of $v$ .

Complete step by step solution:
To launch the spaceship out into the free space, we need to give it a minimum speed such that it gets free from the influence of the gravitational field of the earth. We know that this minimum required speed is called the escape velocity, which is given by
${v_e} = \sqrt {2gR}$ ………………….(1)
To provide the spaceship this much speed, the corresponding kinetic energy required is given by
$K = \dfrac{1}{2}m{v^2}$
Substituting $v = {v_e}$ from (1) we get
$K = \dfrac{1}{2}m{\left( {\sqrt {2gR} } \right)^2}$
On simplifying the above expression, we have
$K = mgR$ ………………….(2)
According to the question, we have
$g = 10m/{s^2}$ ,
$m = 1000kg$ , and
$R = 6400km$
Converting into meters we have
$R = 6.4 \times {10^6}m$
Substituting the above values in (2), we get
$K = 1000 \times 10 \times 6.4 \times {10^6}$
$K = 6.4 \times {10^{10}}J$
Now, the energy required to launch the satellite is equal to the kinetic energy of the spaceship moving with the escape speed. So, the energy required is
$E = K = 6.4 \times {10^{10}}J$
Hence, the correct answer is option D.

Note:
We must note that we have to apply the formula for the escape velocity carefully. We must not forget the factor of $\sqrt 2$ which comes in its expression. If there is any confusion regarding the formula, then we can very easily derive it by equating the kinetic energy of the spaceship with the gravitational potential energy at the surface of the earth.