Question 32: Gravitation

Question

The height vertically above the earth's surface at which the acceleration due to gravity becomes $$1 \%$$ of its value at the surface is

Accepted Answer

Acceleration due to gravity at a height $$h$$ from the earth's surface,
$$\begin{aligned}
g^{\prime} & =\frac{G M}{(R+h)^2} \
	ext { Given, } \quad g^{\prime} & =1 \% 	ext { of } g=\frac{g}{100} \
\Rightarrow \quad \frac{g}{100} & =\frac{G M}{(R+h)^2} \
\frac{(R+h)^2}{100} & =\frac{G M}{g} 	ext { or } \frac{(R+h)^2}{100}=R^2 \quad\left(\because g=\frac{G M}{R^2}ight) \
	ext { or } \quad R+h & =10 R \
h & =9 R
\end{aligned}$$
where, $$R$$ is the radius of the earth.

Answer

8R

Answer

9R

Answer

10R

Answer

20R

Gravitation

The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 \%$ of its value at the surface is

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