a satellite circle Jupiter if placed at a height of 7.80x10^6m above the surface of the planet? The mass of Jupiter 1.90 x 10^ 27kg and the radius of Jupiter is 7.14 x10^7m

Answer 1

Let

mass of Jupiter = m₁ = 1.90x10²⁷ kg

radius of orbit, r = radius of Jupiter + height above the surface of the planet

r = 7.14x10⁷ m + 7.80x10⁶ m = 7.92x10⁷ m

The attractive force between two bodies is given by

F = (Gm₁*m₂)/r²

Where m₂ is the mass of the satellite and G is the universal gravitational constant = 6.67x1^-11 Nm²/kg²

Assuming that the orbit is circular, then the attractive force is equal to the centripetal force, F_c = m₂v²/r, that is

F = F_c ===> (Gm₁*m₂)/r² = m₂v²/r. Solving for the speed v, gives

v = √(Gm₁/r) = √((6.67x1^-11)(1.90x10²⁷)/(7.92x10⁷)) m/s = 40,0001.6 m/s ≈ 40km/s as the orbital speed.