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

Byju's Answer

Standard XII

Physics

Satellites

Two satellite...

Question

Two satellites are moving around the earth in circular orbits at height R and 3R respectively, R being the radius of the earth, the ratio of their kinetic energies is

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

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

Open in App

Solution

The correct option is A
2

$r_{1} = R + h_{1} = R + R = 2 R a n d r_{2} = R + h_{2} = R + 3 R = 4 R K i n e t i c e n e r g y \propto \frac{1}{r} ∴ \frac{(K E)_{1}}{(K E)_{2}} = \frac{r_{2}}{r_{1}} = \frac{4 R}{2 R} = \frac{2}{1}$

Suggest Corrections

Similar questions

Two satellites are moving around the earth in circular orbits at height R and 3R respectively, R being the radius of the earth, the ratio of their kinetic energies is

Q. Two satellites

A

and

B

of the same mass are orbiting the earth at altitudes

R

and

3 R

respectively,where

R

is the radius of the earth. Taking their orbits to be circular obtain the ratios of their kinetic and potential energies.

Q. Two satellites,

A

and

B

, have masses

m

and

2 m

respectively.

A

is in a circular orbit of radius

R

, and

B

in a circular orbit of radius

2 R

around the earth. The ratio of their kinetic energies,

\frac{T_{A}}{T_{B}}

, is:

Q. Two satellites,

A

and

B

, have masses

m

and

2 m

respectively.

A

is in circular orbit of radius

R

, and

B

is in a circular orbit of radius

2 R

around the earth. The ratio of their kinetic energies,

T_{A} / T_{B}

is:

A satellite $X$ is launched into a circular orbit of radius $R$ around the earth while a second satellite $Y$ is launched into an orbit of radius $4 R$ . The ratio of the time periods of $Y$ to $X$ is ______.

Join BYJU'S Learning Program

Two satellites are moving around the earth in circular orbits at height R and 3R respectively, R being the radius of the earth, the ratio of their kinetic energies is

The correct option is A 2 r1=R+h1=R+R=2R and r2=R+h2=R+3R=4RKinetic energy ∝ 1r ∴ (KE)1(KE)2=r2r1=4R2R=21

The correct option is A
2

$r_{1} = R + h_{1} = R + R = 2 R a n d r_{2} = R + h_{2} = R + 3 R = 4 R K i n e t i c e n e r g y \propto \frac{1}{r} ∴ \frac{(K E)_{1}}{(K E)_{2}} = \frac{r_{2}}{r_{1}} = \frac{4 R}{2 R} = \frac{2}{1}$