JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 20)
The gravitational potential at a point above the surface of earth is $$-5.12 \times 10^7 \mathrm{~J} / \mathrm{kg}$$ and the acceleration due to gravity at that point is $$6.4 \mathrm{~m} / \mathrm{s}^2$$. Assume that the mean radius of earth to be $$6400 \mathrm{~km}$$. The height of this point above the earth's surface is :
1600 km
1200 km
540 km
1000 km
Explanation
$$-\frac{G M_E}{R_E+h}=-5.12 \times 10^{-7}$$ .... (i)
$$\frac{G M_E}{\left(R_E+h\right)^2}=6.4$$ ..... (ii)
By (i) and (ii)
$$\Rightarrow h=16 \times 10^5 \mathrm{~m}=1600 \mathrm{~km}$$
Comments (0)
