A satellite moves in a circle around the earth. The radius of this circle is equal to one half the radius of the moon's orbit. This satellite completes one revolution in

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

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

Open in App

Solution

The correct option is D

From Kepler's law of period

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3}$

= $2^{\frac{- 3}{2}}$ of one lunar month.

Suggest Corrections

Similar questions

A satellite moves in a circular orbit around earth. The radius of this orbit is one half that of moon's orbit. The satellite completes one revolution in:

Q. A satellite is revolving around earth in a circular orbit. The radius of orbit is half of the radius of theorbit of moon. Satellite will complete one revolution in.

Q. An artificial satellite is revolving round the earth. The radius of its circular orbit is half the orbital radius of moon. The time taken by this satellite in completing one revolution will be

Q. A satellite orbiting the earth in a circular orbit of radius

R

completes one revolution in

3 h

. If orbital radius of geostationary satellite is

36000 k m

, orbital radius of satellite is

Q. A satellite is to revolve around the earth in a circle of radius 8000 km with what speed should this satellite be projected into the orbit? What will be the time period? (Take g = 9.8 m/

s^{2}

and radius of the earth = 6400 km)

Join BYJU'S Learning Program

A satellite moves in a circle around the earth. The radius of this circle is equal to one half the radius of the moon's orbit. This satellite completes one revolution in

The correct option is D From Kepler's law of period (T1T2)2=(R1R2)3 = 2−32 of one lunar month.

The correct option is D

From Kepler's law of period

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3}$

= $2^{\frac{- 3}{2}}$ of one lunar month.