Question 34: Gravitation

Question

The height at which the acceleration due to gravity becomes $$\frac{g}{16}$$ (where, g = acceleration due to gravity on the surface of the earth) in terms of R is, if R is the radius of earth.

Accepted Answer

As, acceleration due to gravity at height $$h$$,
$$\begin{aligned}
& g^{\prime}=\frac{G M}{(R+h)^2}, \
& \Rightarrow \quad \frac{g}{16}=\frac{G M}{R^2}\left(\frac{R^2}{(R+h)^2}ight)=g \frac{R^2}{(R+h)^2}\left(\because g^{\prime}=\frac{g}{16}ight) \
& \Rightarrow \quad \frac{1}{4}=\frac{R}{R+h} \Rightarrow h=3 R \
\end{aligned}$$

Answer

$$2R$$

Answer

$$3R$$

Answer

$$\sqrt2R$$

Answer

$$\sqrt3R$$

Gravitation

The height at which the acceleration due to gravity becomes $\frac{g}{16}$ (where, g = acceleration due to gravity on the surface of the earth) in terms of R is, if R is the radius of earth.

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