The gravitational potential at a point above the surface of earth is -5.12×10^7 J kg^(-1) and the acceleration due to gravity at that point is 6.4 m…

Question

The gravitational potential at a point above the surface of earth is -5.12×10^7 J kg^(-1) and the acceleration due to gravity at that point is 6.4 m s^(-2). Assume that the mean radius of earth to be 6400 km. The height of this point above the earth's surface is:

MARKS Content Team · Accepted Answer

Correct answer: (A) 1600 km. Gravitational potential at a distance r from the centre of the earth is given by, V=(-GM)/r. Therefore, we can write -GM_E/(R_E+h)=-5.12×10^7 …i also acceleration due to gravity at that point will be (GM_ERE+h^2=6.4 …iiBy (i) and (ii), we get-GM_E)/(R_E+h)(GM_ERE+h^2=5.12×10^7)/6.4 ⇒R_E+h=80×10^5 m=8000 km⇒h=8000 km-6400 km=1600 km

The gravitational potential at a point above the surface of earth is $- 5.12 \times 10^{7} J {kg}^{- 1}$ and the acceleration due to gravity at that point is $6.4 m s^{- 2}$ . Assume that the mean radius of earth to be $6400 km$ . The height of this point above the earth's surface is:

Solution

Relevant Questions

The gravitational potential at a point above the surface of earth is -5.12×107 J kg-1 and the acceleration due to gravity at that point is 6.4 m s-2. Assume that the mean radius of earth to be 6400 km. The height of this point above the earth's surface is:

Solution

Relevant Questions

The gravitational potential at a point above the surface of earth is $- 5.12 \times 10^{7} J {kg}^{- 1}$ and the acceleration due to gravity at that point is $6.4 m s^{- 2}$ . Assume that the mean radius of earth to be $6400 km$ . The height of this point above the earth's surface is: