Question

A body is released at a distance far away from the surface of the earth. Calculate its speed when it is near the surface of earth. Given $g = 9.8\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}$, radius of earth $R = 6.37 \times {10^6}\,{\text{m}}$
A.$22 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}}$
B.$11.2 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}}$
C.$23 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}}$
D.$10 \times {10^3}\,{\text{m}}{{\text{s}}^{ - 1}}$

Answer 1

Find out the initial energy of the body and energy when the body is near the surface of the earth and use the law of conservation of energy to find the velocity or speed of the body. As only the values of $g$ and $R$ are given, try to find the velocity in terms of these two quantities.

Complete Step by step answer:
Given,
$g = 9.8\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}$
Radius of earth $R = 6.37 \times {10^6}\,{\text{m}}$
We will use conservation of energy to find out the velocity of the body.
Let ${(P.E)_{initial}}$ and ${(K.E)_{initial}}$be the initial potential and kinetic energy of the body when the body is released and ${(P.E)_{final}}$ and ${(K.E)_{final}}$ be the final potential and kinetic energy of the body when it comes near to the earth surface.

When the body is released, at that moment its potential and kinetic energy are zero that is,

$${(P.E)_{initial}} = 0 \\\ {(K.E)_{initial}} = 0 \\\$$

When the body comes near the surface of the earth, its potential energy will be
${(P.E)_{final}} = - \dfrac{{GMm}}{R}$
Where $G$is the gravitational constant, $M$is the mass of the earth, $m$is the mass of the body and $R$ is the radius of the earth.
And the kinetic energy of the body is,
${(K.E)_{final}} = \dfrac{1}{2}m{v^2}$
Where $v$ is the velocity of the body.

Now, from conservation of energy we have,
${(P.E)_{initial}} + {(K.E)_{initial}} = {(P.E)_{final}} + {(K.E)_{final}}$
Putting the values of ${(P.E)_{initial}}$, ${(K.E)_{initial}}$, ${(P.E)_{final}}$,

{(K.E)_{final}}$$ we have $$0 + 0 = - \dfrac{{GMm}}{R} + \dfrac{1}{2}m{v^2} \\\ \Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{GMm}}{R} \\\ \Rightarrow v = \sqrt {\dfrac{{2GM}}{R}} \\\

Multiplying and dividing the equation by $\sqrt R$ on R.H.S we get
$v = \sqrt {\dfrac{{2GMR}}{{{R^2}}}}$.................. (i)
We know acceleration due to gravity, $g = \dfrac{{GM}}{{{R^2}}}$
Substituting the $g$in equation (i), we get
$v = \sqrt {2gR}$.................................................. (ii)
Now, putting the values of $g$ and$R$, we get

$$v = \sqrt {2 \times 9.8 \times 6.37 \times {{10}^6}} \\\ \Rightarrow v = 11.17 \times {10^3}\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \\\ \therefore v \sim 11.2 \times {10^3}\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}} \\\$$

Therefore, the speed of the body near the surface is $11.2 \times {10^3}\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$

Hence, the correct answer is option (B) $11.2 \times {10^3}\,{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$

Note: The initial potential and kinetic energy of the body here is zero, as initially the body was at rest. But if the body would be in motion then we would need to consider the potential and kinetic energy of the body as there would be some kinetic energy and might be some potential too.