A remote-sensing satellite of earth revolves in a circular orbit at a height of 0-25 - 106 m above the surface of earth- If earth-s radius is 6-4 - 106 m and g -9-8 ms -2- then the orbital speed of the satellite isA- 6-67 kms -1B- 7-76 kms -1C- 8-56 kms -1D- 9-13 kms -1

The orbital velocity of the satellite is v=√(GMR+h) nbsp; where M= mass of earth and R= radius of earth. so, v=√(6.67× 10^ 11× (6× 10^24)(6.4× 10^6)+(0.25× 10 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Centripetal Force

A remote-sens...

Question

A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^{6}$ $m$ above the surface of earth. If earth's radius is $6.4 \times 10^{6} m$ and $g = 9.8 m s^{- 2}$ , then the orbital speed of the satellite is

6.67 k m s^{- 1}

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

7.76 k m s^{- 1}

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

8.56 k m s^{- 1}

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

9.13 k m s^{- 1}

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

Open in App

Solution

The correct option is B $7.76 k m s^{- 1}$
The orbital velocity of the satellite is $v = \sqrt{\frac{G M}{R + h}}$ where $M =$ mass of earth and $R =$ radius of earth.
so, $v = \sqrt{\frac{6.67 \times 10^{- 11} \times (6 \times 10^{24})}{(6.4 \times 10^{6}) + (0.25 \times 10^{6})}} = 7.76 k m / s$

Suggest Corrections

Similar questions

Q. A remote - sensing satellite of earth revolves in a circular orbit at a height of

0.25 \times 10^{6}

m above the surface of earth. If earth's radius is

6.38 \times 10^{6} m

and

g = 9.8 m s^{- 2}

, then the orbital speed of the satellite is:

Q. A remote-sensing satellite of earth revolves in a circular orbit at a height of

0.25 \times 10^{6}

m

above the surface of earth. If earth's radius is

6.4 \times 10^{6} m

and

g = 9.8 m s^{- 2}

, then the orbital speed of the satellite is

A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 ×10²⁴ kg; radius of the earth = 6.4 ×10⁶ m; G = 6.67 × 10^–11 N m² kg^–2.

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)

Join BYJU'S Learning Program

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25×106 m above the surface of earth. If earth's radius is 6.4×106 m and g=9.8 ms−2, then the orbital speed of the satellite is

The correct option is B 7.76kms−1The orbital velocity of the satellite is v=√GMR+h where M= mass of earth and R= radius of earth. so, v= ⎷6.67×10−11×(6×1024)(6.4×106)+(0.25×106)=7.76 km/s

A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25 \times 10^{6}$ $m$ above the surface of earth. If earth's radius is $6.4 \times 10^{6} m$ and $g = 9.8 m s^{- 2}$ , then the orbital speed of the satellite is

The correct option is B $7.76 k m s^{- 1}$
The orbital velocity of the satellite is $v = \sqrt{\frac{G M}{R + h}}$ where $M =$ mass of earth and $R =$ radius of earth.
so, $v = \sqrt{\frac{6.67 \times 10^{- 11} \times (6 \times 10^{24})}{(6.4 \times 10^{6}) + (0.25 \times 10^{6})}} = 7.76 k m / s$