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Question: An asteroid is moving directly towards the centre of the earth. When at a distance of \(10\)R (R is ...

An asteroid is moving directly towards the centre of the earth. When at a distance of 1010R (R is the radius of the earth) from the earth centre, it has a speed of 1212km/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 11.211.2km/s)? Give your answer to the nearest integer in Kilometre/s_ _ _ _ _ _

Explanation

Solution

As the asteroid is moving towards earth, then by conservation by mechanical energy, the total energy at a distance 10R10R 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=12mv2K.E = \dfrac{1}{2}m{v^2}
3. Potential Energy =−GMmR = \dfrac{{ - GMm}}{R}

Complete step by step answer:
Let the velocity of asteroid at distance 10R10R from centre of Earth by, u=12km/su = 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
=12mu2+(−GMm10R)= \dfrac{1}{2}m{u^2} + \left( {\dfrac{{ - GMm}}{{10R}}} \right)
And Final Mechanical energy == Final kinetic energy ++ initial potential energy
=12mu2+(−GMm10R)= \dfrac{1}{2}m{u^2} + \left( {\dfrac{{ - 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
⇒12mu2+(−GMm10R)=12mv2+(−GMmR)\Rightarrow \dfrac{1}{2}m{u^2} + \left( {\dfrac{{ - GMm}}{{10R}}} \right) = \dfrac{1}{2}m{v^2} + \left( {\dfrac{{ - GMm}}{R}} \right)
⇒12mv2−12mu2=−GMm10R+−GMmR\Rightarrow \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2} = \dfrac{{ - GMm}}{{10R}} + \dfrac{{ - GMm}}{R}
⇒12(v2−u2)=−GMmR(110−1)\Rightarrow \dfrac{1}{2}\left( {{v^2} - {u^2}} \right) = \dfrac{{ - GMm}}{R}\left( {\dfrac{1}{{10}} - 1} \right)
⇒12(v2−u2)=−GMR×−910\Rightarrow \dfrac{1}{2}\left( {{v^2} - {u^2}} \right) = \dfrac{{ - GM}}{R} \times \dfrac{{ - 9}}{{10}}
⇒v2−u2=9GM10R×2\Rightarrow {v^2} - {u^2} = \dfrac{{9GM}}{{10R}} \times 2
⇒v2=u2+9GM5R........(1)\Rightarrow {v^2} = {u^2} + \dfrac{{9GM}}{{5R}}........\left( 1 \right)
Now, here u=12km/s=12000m/su = 12km/s = 12000m/s
M== mass of earth
As GMR2=g⇒GMR=gR\dfrac{{GM}}{{{R^2}}} = g \Rightarrow \dfrac{{GM}}{R} = gR
So, v2=u2+95gR{v^2} = {u^2} + \dfrac{9}{5}gR
⇒v2=(12000)2+95×9.8×6400000\Rightarrow {v^2} = {\left( {12000} \right)^2} + \dfrac{9}{5} \times 9.8 \times 6400000
⇒v2=14000000+112896000\Rightarrow {v^2} = 14000000 + 112896000
⇒v2=256,896,000\Rightarrow {v^2} = 256,896,000
⇒v=16,027.97m/s\Rightarrow v = 16,027.97m/s
⇒v≃16km/s\Rightarrow v \simeq 16km/s

Note:
The relation between acceleration due to gravity, g and universal gravitational constant, G is
g=GMR2⇒GMR=gRg = \dfrac{{GM}}{{{R^2}}} \Rightarrow \dfrac{{GM}}{R} = gR
So, we have replaced GMR\dfrac{{GM}}{R} with gR in equation ....(1).