A satellite of mass $m$ is at a distance $a$ from $a$ star of mass $M$. The speed of satellite is $u$. Suppose the law of universal gravity is $F = - G\frac{{Mm}}{{{r^{2.1}}}}$ instead of $F = - G\frac{{Mm}}{{{r^2}}}$, find the speed of the statellite when it is at $a$ distance $b$ from the star.
$\sqrt {{u^2} + 2GM\left( {\frac{1}{{{b^{1.1}}}} - \frac{1}{{{a^{1.1}}}}} \right)} $
$\sqrt {{u^2} + GM\left( {\frac{1}{{{a^{1.1}}}} - \frac{1}{{{b^{1.1}}}}} \right)}$
$\sqrt {{u^2} + \frac{2}{{1.1}}GM\left( {\frac{1}{{{b^{1.1}}}} - \frac{1}{{{a^{1.1}}}}} \right)}$
$\sqrt {{u^2} + \frac{2}{{2.1}}GM\left( {\frac{1}{{{b^{1.1}}}} - \frac{1}{{{a^{1.1}}}}} \right)}$
Two particles of equal mass go round a circle of radius $R$ under the action of their mutual gravitational attraction. The speed of each particle is
Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ respectively are released in free space with initial separation between their centres equal to $12\,R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is
Masses and radii of earth and moon are $M_1,\, M_2$ and $R_1,\, R_2$ respectively. The distance between their centre is $'d'$ . The minimum velocity given to mass $'M'$ from the mid point of line joining their centre so that it will escape
When a body is taken from pole to the equator its weight
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to