Question

Derive the time period of the satellite orbiting the Earth.

Answer 1

Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to 2π (R_E + h) and time taken for it is the time period, T. Then,

Speed, v = `"Distance travelled"/"Time taken" = (2π ("R"_"E" + "h"))/"T"` ..........(1)

Speed of the satellite, V = `sqrt("GM"_"E"/(("R"_"E" + "h")))`

`sqrt("GM"_"E"/(("R"_"E" + "h"))) = (2π ("R"_"E" + "h"))/"T"`

T = `(2π)/sqrt("GM"_"E") ("R"_"E" + "h")^(3/2)` ............(2)

Squaring both sides of the equation (2) we get,

T² = `(4π^2)/"GM"_"E" ("R"_"E" + "h")^3`

`(4π^2)/"GM"_"E"` = constant say c

T² = c(R_E + h)³

Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R_E. Then,

T² = `(4π^2)/"GM"_"E" "R"_"E"^3`

T² = `(4π^2)/(("GM"_"E"/"R"_"E"^2)) "R"_"E"`

T² = `(4π^2)/"g" "R"_"E"`

Since `"GM"_"E"/"R"_"E"^2` = g

T = `2π sqrt("R"_"E"/"g")`

By substituting the values of R_E = 6.4 × 10⁶ m and g = 9.8 ms⁻², the orbital time period is obtained as T ≅ 85 minutes.