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

F_e = F_s

i.e.

solving and canceling terms we get

Or we can say

We are given

Mass of the sun, M_s = 2×10³⁰ kg

Mass of the earth, M_e = 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