Question 421: Gravitation

Question

The height at which the weight of a body becomes  $${\left( {{1 \over {16}}} \right)^{th}}$$, its weight on the surface of earth (radius R), is

Accepted Answer

Let at h height, the weight of a body becomes 1/16th of its weight on the surface.
$${W_h} = 16{W_s}$$   ...(i)
$$ \Rightarrow $$ $$16{W_s} = {W_s}$$   ...(ii)
$${W_h} = mg = {{GMm} \over {{{\left( {{R_e} + h} \right)}^2}}}$$
$$g' = {{GM} \over {{{\left( {{R_e} + h} \right)}^2}}}$$
Similarly, $$g = {{GM} \over {R_e^2}}$$
$${{g'} \over g} = {{R_e^2} \over {{{\left( {{R_e} + h} \right)}^2}}} \Rightarrow g' = g{\left( {1 + {h \over R}} \right)^{ - 2}}$$
$$ \Rightarrow {W_h} = 16{W_s}$$
$$ \Rightarrow g' = {g \over {16}}$$
$${1 \over {16}} = {\left( {1 + {h \over R}} \right)^{ - 2}} \Rightarrow 4 = 1 + {h \over R}$$
h = 3R

Answer

5R

Answer

15R

Answer

3R

Answer

4R

Gravitation

The height at which the weight of a body becomes ${\left( {{1 \over {16}}} \right)^{th}}$, its weight on the surface of earth (radius R), is

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