The angular speed of earth in $rad/s$, so that bodies on equator may appear weightless is : [Use $g = 10\, m/s^2$ and the radius of earth $= 6.4 \times 10^3\, km$]
$1.25 \times {10^{ - 3}}$
$1.56 \times {10^{ - 3}}$
$1.25 \times {10^{ - 1}}$
$1.56$
A satellite is launched into a circular orbit of radius $R$ around earth, while a second satellite is launched into a circular orbit of radius $1.02\, {R}$. The percentage difference in the time periods of the two satellites is -
If potential energy of a body of mass $m$ on the surface of earth is taken as zero then its potential energy at height $h$ above the surface of earth is [ $R$ is radius of earth and $M$ is mass of earth]
Figure shows the orbit of a planet $P$ round the sun $S.$ $AB$ and $CD$ are the minor and major axes of the ellipse.
If $t_1$ is the time taken by the planet to travel along $ACB$ and $t_2$ the time along $BDA,$ then
A geostationary satellite is orbiting the earth at a height of $6\, R$ from the earth’s surface ($R$ is the earth’s radius ). What is the period of rotation of another satellite at a height of $2.5\, R$ from the earth’s surface
At what height above the earth's surface is the value of $'g'$ is same as in a $200\, km$ deep mine ........ $km$