Question

Periodic time (T) of a satellite revolving above Earth’s surface at a height equal to the radius of Earth (R), is
$g=9.8m/s^2$

• A. $2\pi \sqrt{\frac{2R}{g}}$
• B. $4\sqrt{2}\pi \sqrt{\frac{R}{g}}$
• C. $2\pi \sqrt{\frac{R}{g}}$
• D. $8\pi \sqrt{\frac{R}{g}}$

Answer 1

The correct option is B $4\sqrt{2}\pi \sqrt{\frac{R}{g}}$

Given:
height h = R (radius of earth)
according to the formula for time period
$T=2\pi \sqrt{\frac{(R+h{)}^3}{GM}}$
Since h = R (given), $andg=\frac{GM}{R^2}\Rightarrow GM=g{R}^2$,
$T=2\pi \sqrt{\frac{(R+R{)}^3}{g{R}^2}}$
$T=2\pi \sqrt{\frac{8R}{g}}=4\sqrt{2}\pi \sqrt{\frac{R}{g}}$