The height at which the weight of a body becomes $\frac{1}{16}^{th}$ of its weight on the surface of the Earth (radius $R$) is: (in $R$)

  • A
    $5$
  • B
    $15$
  • C
    $3$
  • D
    $4$

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The value of acceleration due to gravity at a height of $10 \,km$ from the surface of the Earth is $x$. At what depth inside the Earth is the value of the acceleration due to gravity the same value $x$ (in $\,km$)?

What should be the angular velocity of the Earth due to rotation about its own axis so that the weight of an object at the equator becomes $3/5$ of its initial value? (Radius of Earth at the equator is $R = 6400 \, km$,take $g = 10 \, m/s^2$)

The weight of a body at the surface of earth is $18 \, N$. The weight of the body at an altitude of $3200 \, km$ above the earth's surface is $........ \, N$ (given,radius of earth $R_e = 6400 \, km$).

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The force of gravity is least at:

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