If gravitational attraction between two points masses be given by $F=G \frac{m_1 m_2}{r^n}$, then the period of a satellite in a circular orbit will be proportional to

The distance of the centres of Moon and the Earth is $D$. The mass of the Earth is 81 times the mass of the Moon. At what distance from the centre of the Earth, the gravitational force on a particle will be zero?

The gravitational field in a region is given by $$\mathbf{E}=5 \mathrm{~N} / \mathrm{kg} \hat{\mathbf{i}}+12 \mathrm{~N} / \mathrm{kg} \hat{\mathbf{j}}$$. The change in the gravitational potential energy of a particle of mass $$1 \mathrm{~kg}$$ when it is taken from the origin to a point ($$5 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}$$) is

A geostationary satellite is orbiting the Earth at a height of $$4 R$$ above that surface of the Earth. $R$ being the radius of the earth. The, time period of another stellite in coins at a height of $$2 R$$ from the surface of the Earth is.

A skylab or mass $$m \mathrm{~kg}$$ is first launched from the surface of the earth in a circular orbit of radius $$2 R$$ (from the centre of the earth) and then it is shifted from this circular orbit to another circular orbit of radius $$3 R$$. The minimum energy required to place the lab in the first orbit and to shift the lab from first orbit to the second orbit are