Question 347: Gravitation

Question

When the value of acceleration due to gravity '$$g$$' becomes $$\frac{g}{3}$$ above surface of height '$$h$$' then relation between '$$h$$' and '$$R$$' is 
( $$\mathrm{R}=$$ radius of earth)

Accepted Answer

$$\begin{aligned} & \mathrm{g}^{\prime}=\mathrm{g} \frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \quad \mathrm{~g}^{\prime}=\frac{\mathrm{g}}{3} \ & 	herefore \frac{1}{3}=\frac{\mathrm{R}^2}{(\mathrm{R}+\mathrm{h})^2} \quad 	herefore \frac{1}{\sqrt{3}}=\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}} \ & 	herefore \sqrt{3} \mathrm{R}=\mathrm{R}+\mathrm{h} \ & 	herefore \mathrm{h}=(\sqrt{3}-1) \mathrm{R}\end{aligned}$$

Answer

$$\mathrm{h}=\frac{\mathrm{R}}{\sqrt{3}-1}$$

Answer

$$\mathrm{h}=\frac{\sqrt{3}}{\mathrm{R}}$$

Answer

$$h=(\sqrt{2}-1) R$$

Answer

$$\mathrm{h}=(\sqrt{3}-1) \mathrm{R}$$

Gravitation

When the value of acceleration due to gravity '$g' becomes \frac{g}{3} above surface of height 'h' then relation between 'h' and 'R' is ( \mathrm{R}=$ radius of earth)

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