1.

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. What is the time period of another satellite orbiting at a height of 2.5 R from the surface of the earth ?

Answer»

6.2 hour
8.48 hour
9.5 hour
11.6 hour

SOLUTION :`T^(2)PROP R^(3)` (Kepler's law)
`therefore (T_(2))/(T_(1))=((R_(2))/(R_(1)))^(3//2)`
where `R_(2)=2.5R+R=3.5R` and `R_(1)=7R`
`therefore T_(2)=((3.5R)/(7R))^(3//2)xx T_(2)=((1)/(2))^(3//2)xx 24=8.48` hour


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