Simple isn't it? #9

Here's the solution :

By energy conservation we have :

$\displaystyle \frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { Gm{ m }_{ earth } }{ x } =-\frac { 4Gm{ m }_{ earth } }{ 5R }$

$\displaystyle \Rightarrow v = \sqrt { \frac { 2G{ m }_{ e } }{ 5Rx } (5R-4x) }$

To find out $T$ , we have :

$\displaystyle T\sqrt { \frac { 2G{ m }_{ e } }{ 5R } } =\int _{ R }^{ 5R/4 }{ \sqrt { \frac { x }{ 5R-4x } } dx }$

$\displaystyle T=\sqrt { \frac { 5{ R }^{ 3 } }{ 2G{ m }_{ e } } } \int _{ 1 }^{ 5/4 }{ \sqrt { \frac { x }{ 5-4x } } dx }$

Now the integral is equal to $\displaystyle =\dfrac{1}{16}(5\pi+4-10{\tan}^{-1}(2))= 0.53978$

Putting the values we have :

$\displaystyle T = 690.534 s$