Question 265: Gravitation

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 earths
centre, it has a speed of 12 km/s. Neglecting
the effect of earths 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 kilometer/s _____.

Accepted Answer

U
1
+ K
1
= U
2
+ K
2
$$ \Rightarrow $$ $$ - {{GMm} \over {10R}} + {1 \over 2}mV_1^2$$ = $$ - {{GMm} \over R} + {1 \over 2}mV_2^2$$
$$ \Rightarrow $$ $${1 \over 2}V_2^2 = {1 \over 2}V_1^2 + {{GM} \over R} - {{GM} \over {10R}}$$
$$ \Rightarrow $$ $$V_2^2 = V_1^2 + {9 \over 5}{{GM} \over R}$$ ....(1)
Given escape velocity V
e
= 11.2 km/s
$$ \Rightarrow $$ $$\sqrt {{{2GM} \over R}} $$ = 11.2
$$ \Rightarrow $$ $${{{GM} \over R} = {{{{\left( {11.2} ight)}^2}} \over 2}}$$
So from (1)
$$V_2^2 = V_1^2 + {9 \over 5}$$$${ 	imes {{{{\left( {11.2} ight)}^2}} \over 2}}$$
= $${\left( {12} ight)^2}$$ + 112.896
$$ \Rightarrow $$ V
2
= 16 km/s

Gravitation

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