Question asked by Filo student

Change in the value of acceleration due to gravity. If we move $h$ height from the surface of earth vertically upward, is ( $h\ll R$ ) (R-radius of earth, $g$-acceleration due to gravity on 34. If the gravitational force is inversely proportional to surface of earth) ${\left(\frac{7}{2}\right)}^{\text{tn }}$ power of distance $(r)$ then the orbital velocity

1. $\frac{2h}{R}$
2. $\frac{2h}{gR}$ is proportional to
3. $r^{-34})$
4. $r^2$
5. $\frac{2gh}{R}$
6. $\frac{2g}{Rh}$
7. $r^{-34})$
8. $r^{2\pi }$

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