Question

An asteroid is moving directly towards the centre of the earth. When at a distance of $10$R (R is the radius of the earth) from the earth centre, it has a speed of $12$km/s. Neglecting the effect on the earth's atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is $11.2$km/s)? Give your answer to the nearest integer in Kilometre/s_ _ _ _ _ _

Answer 1

Hint: As the asteroid is moving towards earth, then by conservation by mechanical energy, the total energy at a distance $10R$ from the centre of earth will be the same as the total energy at distance R from the centre of earth.
Formula used:
1. Initial Mechanical Energy $=$ Final Mechanical Energy
2. $K.E={\displaystyle \frac{1}{2}}m{v}^2$
3. Potential Energy $={\displaystyle \frac{-GMm}{R}}$

Complete step by step answer:
Let the velocity of asteroid at distance $10R$ from centre of Earth by, $u=12km/s$ and the final velocity of asteroid at a distance R from the centre of earth be v km/s as shown in figure below:

Now, initial Mechanical Energy $=$ Initial kinetic energy $+$ initial potential energy
$={\displaystyle \frac{1}{2}}m{u}^2+\left({\displaystyle \frac{-GMm}{10R}}\right)$
And Final Mechanical energy $=$ Final kinetic energy $+$ initial potential energy
$={\displaystyle \frac{1}{2}}m{u}^2+\left({\displaystyle \frac{-GMm}{10R}}\right)$
Where M is mass of Earth
m is mass of asteroid
R is radius of earth and
G is universal Gravitational constant
So, by conservation of mechanical energy, initial mechanical energy $=$ final mechanical energy
$\Rightarrow {\displaystyle \frac{1}{2}}m{u}^2+\left({\displaystyle \frac{-GMm}{10R}}\right)={\displaystyle \frac{1}{2}}m{v}^2+\left({\displaystyle \frac{-GMm}{R}}\right)$
$\Rightarrow {\displaystyle \frac{1}{2}}m{v}^2-{\displaystyle \frac{1}{2}}m{u}^2={\displaystyle \frac{-GMm}{10R}}+{\displaystyle \frac{-GMm}{R}}$
$\Rightarrow {\displaystyle \frac{1}{2}}\left(v^2-u^2\right)={\displaystyle \frac{-GMm}{R}}\left({\displaystyle \frac{1}{10}}-1\right)$
$\Rightarrow {\displaystyle \frac{1}{2}}\left(v^2-u^2\right)={\displaystyle \frac{-GM}{R}}\times {\displaystyle \frac{-9}{10}}$
$\Rightarrow v^2-u^2={\displaystyle \frac{9GM}{10R}}\times 2$
$\Rightarrow v^2=u^2+{\displaystyle \frac{9GM}{5R}}........\left(1\right)$
Now, here $u=12km/s=12000m/s$
M$=$ mass of earth
As ${\displaystyle \frac{GM}{R^2}}=g\Rightarrow {\displaystyle \frac{GM}{R}}=gR$
So, $$v^2=u^2+{\displaystyle \frac{9}{5}}gR$$
$$\Rightarrow v^2={\left(12000\right)}^2+{\displaystyle \frac{9}{5}}\times 9.8\times 6400000$$
$$\Rightarrow v^2=14000000+112896000$$
$$\Rightarrow v^2=256,896,000$$
$$\Rightarrow v=16,027.97m/s$$
$\Rightarrow v\simeq 16km/s$

Note:
The relation between acceleration due to gravity, g and universal gravitational constant, G is
$g={\displaystyle \frac{GM}{R^2}}\Rightarrow {\displaystyle \frac{GM}{R}}=gR$
So, we have replaced ${\displaystyle \frac{GM}{R}}$ with gR in equation ....(1).