Question

Gravitation question?

a satellite P revolving around the earth at a height h=radius of the earth above the equator. another satellite Q is at a height of 2h revolving in opposite direction. At an instant, the two are at same vertical line passing through the centre of sphere.find the least time of after which again they are in this situation.

Answer 1

After `43` minutes and `46` seconds.

Explanation:

Speed of a satellite around Earth, which is at a height of `h=R`, the radius of Earth i.e. at a distance of `2R` from Earth's center, is given by `v=sqrt((GM)/(2R))`

As the distance covered is `2pixx2R=4piR`, time taken by it is given by

`T_1=(4piR)/sqrt((GM)/(2R))=4pisqrt((2R^3)/(GM)`

For a satellite at a height of `h=2R`, the radius of Earth i.e. at a distance of `3R` from Earth's center, is given by `v=sqrt((GM)/(3R))` and as distance would be `6piR`, time taken would be

`T_2=(6piR)/sqrt((GM)/(3R))=6pisqrt((3R^3)/(GM)`

and `T_1/T_2=(2/3)^(3/2)=0.5443`

Hence the two satellites will be again at the same vertical line passing through the center of Earth after `1/0.5443~=1.837` revolutions of the first (lower) satellite.

or `1.837xx4pisqrt((2R^3)/(GM)`

= `1.837xx4pisqrt((2xx(6.367xx10^6)^3)/(6.67xx10^(-11)xx5.98xx10^24)`

= `1.837xx4xx3.14159xx10xxsqrt((20xx6.367^3)/(6.67xx5.98))`

= `1.837xx4xx3.14159xx10xx11.376`

= `2626.08` seconds

or `43` minutes and `46` seconds.