A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours.

Explanation

As `T^(2)prop R^(3)` …. (By Kepler's third law)
`therefore (T_(1)^(2))/(T_(2)^(2))=(R_(1)^(3))/(R_(2)^(3))`
or `((T_(2))/(T_(1)))^(2)=((R_(2))/(R_(1)))^(3)`
`rArr T_(2)^(2)=T_(1)^(2)[(3.5R)/(7R)]^(3)`
`rArr T_(2)^(2)=(T_(1)^(2))/(8)`
`rArr T_(2)=(T_(1))/(2sqrt(2))`
`T_(2)=(24)/(2sqrt(2)) " " ....(because T_(1)=24 "hours")`
`therefore T_(2)=6sqrt(2)` hours.