A rocket is fired from the earth towards the sun. At what distance from the earth’s
centre is the gravitational force on the rocket zero ? Mass of the sun = 2×1030kg, mass of the
earth = 6×1024kg. Neglect the effect of other planets etc. (orbital radius =1.5 ×1011m).
urgent...plss help...it is 10 raise to 24 and 1o raise to 30
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Mass of the Sun (Ms) = 2 × 1030 kg
Mass of the Earth (Me) = 6 × 10 24 kg
Orbital radius, r = 1.5 × 1011 m
Mass of the rocket = m
Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.
From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:
GmMs / (r - x )^2 = GmMe / x^2
[ ( r - x ) / x ]^2 = Ms / Me
( r - x ) / x = [ 2 × 10^30 / 60 × 10^24]^1/2
= 577.35
1.5 × 10^11 - x = 577.35x
578.35 x = 1.5 × 10^11
x = 1.5 × 10^11 / 578.35 = 2.59 × 10^8 m.
Mass of the Earth (Me) = 6 × 10 24 kg
Orbital radius, r = 1.5 × 1011 m
Mass of the rocket = m
Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.
From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:
GmMs / (r - x )^2 = GmMe / x^2
[ ( r - x ) / x ]^2 = Ms / Me
( r - x ) / x = [ 2 × 10^30 / 60 × 10^24]^1/2
= 577.35
1.5 × 10^11 - x = 577.35x
578.35 x = 1.5 × 10^11
x = 1.5 × 10^11 / 578.35 = 2.59 × 10^8 m.
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