The weight of an object on the surface of the Earth is $72 \, N$. What will be the weight of the object at a height of $R/2$ from the surface of the Earth? ($R$ = radius of the Earth)
- A$28$
- B$16$
- C$32$
- D$72$
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At a height $R$ above the earth's surface,the gravitational acceleration is ($R$ = radius of earth,$g$ = acceleration due to gravity on earth's surface).
Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$ $(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth $(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles $(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator
| Column-$I$ | Column-$II$ |
| $(1)$ Maximum value of acceleration due to gravity $g$ | $(a)$ At the center of the Earth |
| $(2)$ Minimum value of acceleration due to gravity $g$ | $(b)$ At the poles |
| $(3)$ Zero value of acceleration due to gravity $g$ | $(c)$ At the equator |
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View Solution$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force on it at a height equal to half the radius of the earth (in $N$)?
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View SolutionThe value of acceleration due to gravity $g$ is maximum at
At what depth $d$ below the surface of the Earth does the acceleration due to gravity become $\frac{1}{n}$ times its value at the surface? ($R$ = radius of the Earth)
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