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]

$2 π \sqrt{\frac{2 R}{g}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$4 \sqrt{2} π \sqrt{\frac{R}{g}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$2 π \sqrt{\frac{R}{g}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$8 π \sqrt{\frac{R}{g}}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$8 π \sqrt{\frac{R}{g}}$

$T = 2 π \sqrt{\frac{(R + h)^{3}}{G M}} = 2 π \sqrt{\frac{(R + R)^{3}}{g R^{2}}} = 2 π \sqrt{\frac{8 R}{g}} = 8 π \sqrt{\frac{R}{g}} [A s h = R (g i v e n)]$

Suggest Corrections

Similar questions

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]

Q. Time period of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is

Q. A geostationary satellite is orbiting the earth at a height

6 R

above the surface of the earth, where

R

is the radius of earth. The time period of another satellite revolving around earth at a height

2.5 R

from earth's surface is :

A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes $5.26 \times 10^{3}$ seconds to complete a revolution while its centripetal acceleration is $9.92 m / s^{2}$ . Height of satellite above surface of earth is (Radius of earth $6.37 \times 10^{6}$ m)

Q. A satellite of mass m is revolving round the earth at a height R above the surface of earth. If g is gravitational acceleration at surface of earth and R is radius of earth then magnitude of momentum of satellite is

Join BYJU'S Learning Program

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]

The correct option is D 8π√Rg T=2π√(R+h)3GM=2π√(R+R)3gR2=2π√8Rg=8π√Rg [As h=R (given)]

The correct option is D
$8 π \sqrt{\frac{R}{g}}$

$T = 2 π \sqrt{\frac{(R + h)^{3}}{G M}} = 2 π \sqrt{\frac{(R + R)^{3}}{g R^{2}}} = 2 π \sqrt{\frac{8 R}{g}} = 8 π \sqrt{\frac{R}{g}} [A s h = R (g i v e n)]$