Question 45: Gravitation

Question

If a body of mass 1 kg falls on the earth from infinity, it attains velocity (
v
) and kinetic energy (
k
) on reaching the surface of earth. The values of
v
and
k
respectively are __________.
(Take radius of earth to be 6400 km and
g
= 9.8 m/s
2
)

Accepted Answer

For a body falling from infinity to the surface of the earth, the gain in kinetic energy is equal to the loss in gravitational potential energy.
For a mass $m$ falling from infinity:
$$ \frac{1}{2}mv^2 = \frac{GMm}{R} $$
Also, we know that
$$ g = \frac{GM}{R^2} \Rightarrow GM = gR^2 $$
Substitute this into the energy equation:
$$ \frac{1}{2}mv^2 = \frac{gR^2 m}{R} = mgR $$
So,
$$ \frac{1}{2}mv^2 = mgR $$
For $m = 1\,	ext{kg}$:
$$ \frac{1}{2}v^2 = gR $$
Given:
$$ R = 6400\,	ext{km} = 6.4 	imes 10^6\,	ext{m} $$
and
$$ g = 9.8\,	ext{m/s}^2 $$
Now,
$$ v = \sqrt{2gR} $$
$$ v = \sqrt{2 	imes 9.8 	imes 6.4 	imes 10^6} $$
$$ v = \sqrt{125.44 	imes 10^6} $$
$$ v = 1.12 	imes 10^4\,	ext{m/s} $$
$$ v = 11.2\,	ext{km/s} $$
Now kinetic energy:
$$ k = \frac{1}{2}mv^2 = mgR $$
$$ k = 1 	imes 9.8 	imes 6.4 	imes 10^6 $$
$$ k = 62.72 	imes 10^6\,	ext{J} = 6.27 	imes 10^7\,	ext{J} $$
So, the correct values are:
$$ v = 11.2\,	ext{km/s}, \qquad k = 6.27 	imes 10^7\,	ext{J} $$
Therefore, the correct option is:
$$ \boxed{	ext{Option A}} $$

Answer

11.2 km/s; $6.27 	imes 10^7$ J

Answer

11.2 km/s; $12.54 	imes 10^7$ J

Answer

8.8 km/s; $6.27 	imes 10^7$ J

Answer

8.8 km/s; $12.54 	imes 10^7$ J

Gravitation

If a body of mass 1 kg falls on the earth from infinity, it attains velocity ( v ) and kinetic energy ( k ) on reaching the surface of earth. The values of v and k respectively are __________. (Take radius of earth to be 6400 km and g = 9.8 m/s 2 )

Choose the correct answer: