Question

Suppose the gravitational force varies inversely as the n^th power of the distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to :

• A. Rⁿ
• B. R^(n+1)/2
• C. R^(n–1)/2
• D. R^–n

Answer 1

Answer: (B)

R^(n+1)/2

Answer 2

T =

E = $\frac { 1 } { 2 }$ mv² = $\left[ \frac { 2 \mathrm { GM } } { \mathrm { R } ^ { \mathrm { n } - 1 } } \right] ^ { 1 / 2 }$

\ T = $\frac { 2 \pi \mathrm { R } } { \sqrt { 2 \mathrm { GM } / \mathrm { R } ^ { \mathrm { n } - 1 } } } = \frac { 2 \pi } { \sqrt { 2 \mathrm { GM } } }$ × R^(n+1)/2

\ T µ R^{(n + 1)/2}.