MyQuestionIcon

1

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

Satellite Launch Vehicle and Jounney to PSLV

What is the d...

Question

What is the distance of geo-stationary satellites from the Earth?

A

37, 956

km

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

B

35, 786

km

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

C

43, 000

km

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

D

23, 123

km

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

Open in App

Solution

The correct option is B $35, 786$ km
Distance of a geostationary satellite from the Earth's surface is kept at $35786 k m$ so that its time period of revolution is $24$ hours.

Suggest Corrections

thumbs-up

0

Similar questions

Q. A geo-stationary satellite orbits around the earth in a circular orbit of radius

36000 k m

. Then, the time period of a spy satellite orbiting a few

100 k m

above the earth's surface (

R_{e a r t h} = 6400 k m

) will approximately be -

Q. A geo-stationary satellite is orbiting around earth at height of

30, 000

km in circular orbit. The radius of the earth is taken as

6000

km. The geo-stationary satellite comes back to its position after one revolution in exactly

24

hours. Let the acceleration due to gravity (g)

10 m / s^{2}

and mass of satellite be

1000

;

calculate the work done in

12

hours when moving under gravitational force.

Q. The angles of elevation of an artificial satellite as measured from two earth stations are

30^{\circ}

60^{\circ}

. If the distance between the earth stations is 4000 km, the height of the satellite is

Q. The ratio of the orbital speeds of two satellites of the earth if the satellites are at heights

6400

19200

km from the surface of the Earth. (Radius of the earth

= 6400

Q. A geo-stationary satellite orbits around the earth in a circular orbit of radius

36, 000 km

. Then, the time period of a spy satellite orbiting a few hundred km above the earth’s surface

(R_{earth} = 6, 400 km)

will approximately be-
(Assume

{(\frac{8}{45})}^{1.5} = 0.08

)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Satellite Launch Vehicle and Jounney to PSLV Tackle

PHYSICS

Watch in App

explore_more_icon

Explore more

Satellite Launch Vehicle and Jounney to PSLV

Standard X Physics

Join BYJU'S Learning Program

CrossIcon