$A$ spaceship orbits around a planet at a height of $20 \, km$ from its surface. Assuming that only the gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in $24 \, hours$ around the planet? [Given: Mass of planet $= 8 \times 10^{22} \, kg$, Radius of planet $= 2 \times 10^6 \, m$, Gravitational constant $G = 6.67 \times 10^{-11} \, N \cdot m^2/kg^2$]

  • A
    $9$
  • B
    $11$
  • C
    $13$
  • D
    $17$

Explore More

Similar Questions

Assertion $A$: An astronaut inside a massive spaceship orbiting around the Earth will experience a finite but small gravitational force.
Reason $R$: The centripetal force necessary to keep the spaceship in orbit around the Earth is provided by the gravitational force between the Earth and the spaceship.

Select the correct statement from the following:

The motion of planets in the solar system is governed by which of the following conservation laws?

$A$ planet is revolving around the sun in an elliptical orbit. Its closest distance from the sun is $r_{min}$,and the farthest distance from the sun is $r_{max}$. If the orbital angular velocity of the planet when it is nearest to the sun is $\omega$,then the orbital angular velocity at the point when it is at the farthest distance from the sun is:

Difficult
View Solution

$A$ planet revolves around the Sun in an elliptical orbit. When it is closest to the Sun,its distance from the Sun is $d_1$ and its velocity is $v_1$. When it is farthest from the Sun,its distance from the Sun is $d_2$. What is its velocity at that point?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo