Question

A geostationary satellite is orbiting the earth at the height of 6R above the surface of earth. R being radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is ____________.

विकल्प

• 10 hour

• `6/sqrt2` hour

• 6 hour

• `6 sqrt2` hour

Answer 1

A geostationary satellite is orbiting the earth at the height of 6R above the surface of earth. R being radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is `6 sqrt2` hour.

Explanation:

From Kepler's III'rd law, we have

`"T"^2 prop "R"^3`

`rightarrow ("T"_2/"T"_1)^2 = ("R"_2/"R"_1)^3`

Here, R2 = 2.5 R + R = 3.5 R, from the centre of earth

R₁ = 6R + R = 7R, from the centre of earth

Therefore, `("T"_2/"T"_1)^2 = ((3.5"R")/(7"R"))^3 = (1/2)^3`

`Rightarrow "T"_2 = "T"_1/(2)^(3/2)`

Time period for a geo-stationary satellite is T₁ =24 hr.

Hence, `"T"_2 = 24/(2)^(3/2) = 24/(2sqrt2)`

` = 6 sqrt2  "hr"`