If the gravitational force between two objects were proportional to $\frac{1}{R}$ (and not as $\frac{1}{R^2}$) where $R$ is the separation between them,then a particle in a circular orbit under such a force would have its orbital speed $v$ proportional to

If the Earth is considered a point mass of $6 \times 10^{24} \ kg$ revolving around the Sun at a distance of $1.5 \times 10^8 \ km$ in a period of $T = 3.14 \times 10^7 \ s$,then the angular momentum of the Earth around the Sun will be:

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

Two point masses of mass $m_1 = fM$ and $m_2 = (1 - f)M$ $(f < 1)$ are in outer space (far from the gravitational influence of other objects) at a distance $R$ from each other. They move in circular orbits about their centre of mass with angular velocities $\omega_1$ for $m_1$ and $\omega_2$ for $m_2$. In that case:

If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth,the height of the satellite above the surface of the earth is

Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$,then ${T^2}$ is proportional to: