Orbital velocity of earth's satellite near the surface is 7 km/s. If the radius of the orbit is 4 times than that of Earth's radius, what will be the orbital velocity in the orbit?

Explanation

The radius of the orbit is 4 times the radius of the earth(R), the orbital speed (v′) is given by: v′ = √GMe/4R = 1/√4 × √GMe/R = 0.5×7 = 3.5 km/s. Given: Orbital velocity near the surface (v₁) = 7 km/s Radius of the new orbit (r) = 4 times the Earth's radius We can use the principle of conservation of angular momentum: Angular momentum (L) = constant L₁ = L₂ (Initial angular momentum = Final angular momentum) For a circular orbit, the angular momentum (L) is given by: L = mvr Where: m = mass of the satellite v = orbital velocity r = radius of the orbit In the original orbit (L₁): m * v₁ * r₁ = constant In the new orbit (L₂): m * v₂ * r₂ = constant Since the mass of the satellite remains constant, we can set L₁ equal to L₂: v₁ * r₁ = v₂ * r₂ We know that r₂ = 4 * r₁ (radius of the new orbit is 4 times the Earth's radius) v₂ = (v₁ * r₁) / r₂ v₂ = (7 km/s * r₁) / (4 * r₁) v₂ = 7 km/s / 4 v₂ = 1.75 km/s So, the correct orbital velocity in the new orbit is 1.75 km/s. The radius of the orbit is 4 times the radius of the earth(R), the orbital speed (v′) is given by: v′ = √GMe4R = 1√4√GMeR = 0.5×7 = 3.5 km/s.