Question

An asteroid is moving directly towards the centre of the Earth. When at a distance of $10R$ ( $R$ is the radius of the Earth ) from the Earth's centre, it has a speed of $12km/s$. Neglecting the effect on 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 $12km/s$ )? Give your answer to the integer in $km/s$.

Answer 1

Step 1. Given data:

The distance of asteroid from the Earth's centre, $d=10Rkm$

The radius of the Earth $=$ $R$

The speed of an asteroid, $v_i=12km/s$

The escape velocity of an asteroid, $v_e=12km/s$

Step 2. Formula used:

By the application of conservation of energy before and after the collision,

Initial [Potential energy $+$ Kinetic energy] $=$Final [Potential energy $+$ Kinetic energy]

i.e. $[U_i+K_i]=[U_f+K_f]$

$-\frac{GMm}{10R}+\frac{1}{2}m{v}_i^2=-\frac{GMm}{R}+\frac{1}{2}m{v}_f^2.......\left(1\right)$

where, $"G"$is the Gravitational constant

$"M"$is the Mass of Earth

$"v_f"$ is the speed of an asteroid while hitting the Earth's surface

Step 3. Finding the speed of an asteroid when it hits the surface of the Earth:

By re-arranging the equation $\left(1\right)$ we get,

$\frac{1}{2}m{v}_f^2=\frac{9MmG}{10R}+\frac{1}{2}m{(12\times {10}^3)}^2[\because 1km=1000m]$ $$\begin{gathered} v_f^2=\frac{9\times 2GM}{10R}+{(12\times {10}^3)}^{2}m/s\Rightarrow v_f=\sqrt{\frac{9{\left(v_e\right)}^2}{10}+{(12\times {10}^3)}^2}m/s[\because v_e=\sqrt{\frac{2GM}{R}}=12km/s] \\ {v}_f=\sqrt{\frac{9{(12\times {10}^3)}^2}{10}+{(12\times {10}^3)}^2}m/s \\ {v}_f=(12\times {10}^3)\sqrt{1.9}=16.5\times {10}^{3}m/s \\ {v}_f=16.5km/s \end{gathered}$$

Hence, the speed of the asteroid when it hits the surface of the Earth is $v_f=16.5km/s$