The height of satellite above earth surface is $36000 k m$ and radius of earth is $6400 k m$ and period of revolution of satellite is $24 h r s$ . The linear speed of the satellite will be

3.1 \times 10^{8} m / s

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

4.1 \times 10^{6} m / s

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

6 \times 10^{8} m / s

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

5.8 \times 10^{6} m / s

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

Open in App

Solution

The correct option is A $3.1 \times 10^{8} m / s$
Radius of orbit of satellite
$= R + h = (6400 + 36000) k m$
$= 42400 k m = 4.24 \times 10^{7} m$
Period of revolution
$= 24 h r s = 24 \times 60 \times 60 = 86400 s e c$
$⟹$ linear speed
$v = \frac{2 π r}{T} = \frac{2 \times 3.14 \times 4.24 \times 10^{7}}{86400} = 3.1 \times 10^{8} m / s$

Suggest Corrections

Similar questions

Q. A satellite is moving in a circular orbit at a certain height above the earth's surface. It takes

5.26 \times 10^{3} s

to complete one revolution with a centripetal acceleration equal to 9.32 m/s

^{2}

. The height of satellite orbiting above the earth is
(Earth's radius

= 6.37 \times 10^{6} m

)

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)