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Question
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 ×1030 kg, mass of the earth = 6 × 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).
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Solution 1
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:
`(GmM_s)/(r-x)^2 = GmM_e/x^2`
((r-x)/x)^2 = `M_s/M_e`
`(r-x)/r = ((2xx10^(30))/(60xx10^(24)))^(1/2) = 577.35`
`1.5 xx 10^(11)-x=577.35x`
`x = (1.5xx10^11)/578.35` = `2.59 xx 10^8` m
Solution 2
Mass of Sun, M = 2 x 1030 kg; Mass of Earth, m = 6 x 1024 kg Distance between Sim and Earth, r = 1.5 x 1011 m
Let at the point P, the gravitational force on the rocket due to Earth = gravitional force on the rocket due to sun
Let x = distance of the point { from the Earth
Then `(Gm)/x^2 = (GM)/(r-x)^2`
`=>(r-x)^2/x^2 = M/m = (2xx10^(30))/(6xx10)^24 = 10^6/3`
or `(r-x)/x = 10^3/sqrt3 => r/x = 10^3/sqrt3 + 1 =~ 10^3/sqrt3`
or `x = (sqrt3r)/10^3 = (1.732xx1.5xx10^11)/10^3 = 2.6 xx 10^8` m
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