Question

Calculate the critical velocity of the satellite to be located at 35780 km above the surface of earth.

Answer 1

Given: Height of the satellite above the earth’s surface (h)

= 35780 km

= 35780 × 10³ m

We know that:

Gravitational constant (G) = 6.67 × 10^–11 N m²/kg²,

mass of earth (M) = 6 × 10²⁴ kg,

radius of earth (R) = 6400 km

= 6400 × 10³ m

To find: Tangential velocity of satellite (v_c)

Formula: `"v"_"c" = sqrt("GM"/("R + h"))`

Calculation: From formula,

`"v"_"c" = sqrt(((6.67 xx 10^-11) xx (6 xx 10^24))/((6400 + 35780) xx 10^3))`

`= sqrt((40.02 xx 10^13)/(42180 xx 10^3))`

`= sqrt(40.02/42180 xx 10^10)`

`= sqrt (0.0009487909 xx 10^10)`

`= sqrt(9.487909 xx10^6)`

≈ `sqrt9.5 xx 10^3`

= 3.08 × 10³ m/s

= 3.08 km/s

The critical velocity of the satellite is 3.08 km/s.