Question

Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [g is acceleration due to gravity at Earth’s surface]

• 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. $4\sqrt{2}\pi \sqrt{\frac{R}{g}}$.

We know that, the time period of a satellite revolving above Earth’s surface at a height equal to h is given by,

$T=2\pi \sqrt{\frac{(R+h{)}^3}{GM}}$

where R is the radius of the earth and M is the mass of the earth.

Here, h = R, therefore,

$T=2\pi \sqrt{\frac{(R+R{)}^3}{g{R}^2}}$

[Since $g=\frac{GM}{R^2}$]

$T=2\pi \sqrt{\frac{8R}{g}}=4\sqrt{2}\pi \sqrt{\frac{R}{g}}$.