A rocket is fired from the earth towards the sun. At what distance from the earth’s center 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).
The net gravitational force on the rocket will be zero when attractive gravitational force on Rocket due to sun is equal in magnitude to attractive gravitational force by Earth, so it is between earth and sun, suppose distance between earth and sun or the orbital radius is R, and rocket is at a distance r from earth so it is at (R-r) distance from sun and attractive pull balances attractive pull of earth
The situation has been depicted in the figure
The gravitational force on Rocket due to sun Fs is equal in magnitude to attractive gravitational force by Earth Fe and opposite in direction as can be seen in the figure
we know gravitational force on a body is given as
Where F is the gravitational force
G is universal gravitational Constant
m1 is mass of the first body
m2 is the mass of the second body
and r is the distance between the two bodies
Now gravitation force on rocket due to earth will be
Where Me and Mr are masses of earth and rocket and r is a separation between them
Similarly, gravitation force on rocket due to Sun will be
Where Ms and Mr are masses of Sun and rocket and separation between Sun and rocket is (R-r)
Since both forces should be equal in magnitude, equating both
Fe = Fs
i.e.
solving and canceling terms we get
Or we can say
We are given
Mass of the sun, Ms = 2×1030 kg
Mass of the earth, Me = 6×1024 kg
The distance between Earth and Sun or orbital Radius
R = 1.5 × 1011 m
So putting these values to find the distance between earth and rocket r
577.3r = 1.5 × 1011 m – r
578.3r = 1.5 × 1011 m
r = (1.5 × 1011 m)/ 578.3 = 2.59 × 103 m
so the rocket is at a distance of 2.59 × 103 m from earth’s centre