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

Byju's Answer

Standard XII

Physics

Satellites

A heavenly bo...

Question

A heavenly body of mass m revolves around a planet of mass M in a circular orbit with a constant speed v. The radius of the orbit is given by

\frac{G M}{v^{2}}

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

\frac{m v^{2}}{G M}

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

\frac{G M}{v^{3}}

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

\frac{v^{2}}{G M}

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

Open in App

Solution

The correct option is A $\frac{G M}{v^{2}}$
$\frac{G M m}{r^{2}} = \frac{m v^{2}}{r}$
$\Rightarrow r = \frac{G M}{v^{2}}$
Hence,
option $(A)$ is correct answer.

Suggest Corrections

Similar questions

Q. If a satellite is revolving around a planet of mass M in an elliptical orbit of semi-major axis a. Show that the orbital speed of the satellite when it is at a distance r from the planet will be given by

V^{2} = G M [\frac{2}{r} - \frac{1}{a}]

Q. A satellite is revolving around earth in a circular orbit of radius

2 R

, where

R

is the earth's radius. A particle is projected from this satellite radially outward relative to satellite with speed

u

. Find the minimum value of

u

so that particle escapes the gravitational force of earth.

M

is the mass of earth. (Mass of earth >>> Mass of satellite >>> Mass of particle)

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is $R_{o}$ and mass of the earth M, the angular momentum about the centre of the earth is

Q. A meteorite approaching a planet of mass

M

(in the straight line passing through the centre of the planet) collides with an automatic space station orbiting the planet in a circular trajectory of radius

R

. The mass of the station is ten times as large as the mass of the meteorite. As a result of the collision, the meteorite sticks in the station which goes over to a new orbit the minimum distance

R / 2

from the planet. Speed of the meteorite just before it collides with the planet is:

Q. A body of mass 100 gm is rotating in a circular path of radius r with constant speed. The work done in one complete revolution is-

Join BYJU'S Learning Program

A heavenly body of mass m revolves around a planet of mass M in a circular orbit with a constant speed v. The radius of the orbit is given by

The correct option is A GMv2GMmr2=mv2r⇒r=GMv2Hence,option (A) is correct answer.

The correct option is A $\frac{G M}{v^{2}}$
$\frac{G M m}{r^{2}} = \frac{m v^{2}}{r}$
$\Rightarrow r = \frac{G M}{v^{2}}$
Hence,
option $(A)$ is correct answer.