A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is____________ × 10¹⁰ J.
(Mass of earth =6 × 10²⁴ kg, Radius of earth =6.4 × 10⁶ m, Gravitational constant =6.67 × 10⁻¹¹ Nm² kg⁻² )

Solution

Physics sir · 4 steps

Gravity is the centripetal force, so K = GMm/2r with r taken from Earth's centre.

1. Gravity bends the path. In a circular orbit, gravity provides the centripetal force: GMm/r² = mv²/r. So mv² = GMm/r.

   K = 1/2mv² = GMm/2r

2. Find the orbit radius. Measure r from Earth's centre, not from its surface: r = R + h = 6.4 × 10⁶ + 0.27 × 10⁶ = 6.67 × 10⁶ m.
3. Put in numbers. K = (6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1000)/(2 × 6.67 × 10⁶). The 6.67 cancels, which leaves K = (6 × 10¹⁶)/(2 × 10⁶) = 3 × 10¹⁰ J.
4. So. The kinetic energy is 3 × 10¹⁰ J, so the blank is 3.

The common slip: 6. Dropping the factor 1/2 and using K = GMm/r gives 6, which is the magnitude of the potential energy, not the kinetic energy.

The idea to keep. For any circular orbit, K = GMm/2r, U = -GMm/r and E = -K. Always take r = R + h, measured from the centre of the Earth.

Answer from the NTA final answer key. Solution by PrepWiser.