If $r$ denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to

Kepler's third law states that square of period of revolution $(T)$ of a planet around the sun, is proportional to third power of average distance $r$ between sun and planet i.e.

$\therefore \;{T^2} = k{r^3}$

here $K$ is constant.

If the masses of sun and planet are $M$ and $m$ respectively then as per Newton's law of gravitation force of attraction between them is $F = \frac{{GMm}}{{{r^2}}}$ , here $G$ gravitational constant . The relation between $G$ and $K$ is described as

The planet Mars has two moons, if one of them has a period $7\, hours,\, 30\, minutes$ and an orbital radius of $9.0 \times 10^{3}\, {km} .$ Find the mass of Mars.

$\left\{\operatorname{Given} \frac{4 \pi^{2}}{G}=6 \times 10^{11} {N}^{-1} {m}^{-2} {kg}^{2}\right\}$

What does not change in the field of central force

In planetary motion the areal velocity of position vector of a planet depends on angular velocity $(\omega )$ and the distance of the planet from sun $(r)$. If so the correct relation for areal velocity is

Remote sensing satellites move in an orbit that is at an average height of about $500 \,km$ from the surface of the earth. The camera onboard one such satellite has a screen of area $A$ on which the images captured by it are formed. If the focal length of the camera lens is $50 \,cm$, then the terrestrial area that can be observed from the satellite is close to ............... $A$