Question

A particle is projected with speed \(4\,km/s\). Find maximum height (in km). Radius of earth \(=6400\,km\), \(g=9.8\,m/s^2\).

Hint:

For heights comparable to Earth's radius (\(h \gtrsim R\)), use \(\frac{1}{2}mv^2 = mgR^2\left(\frac{1}{R} - \frac{1}{R+h}\right)\).

Answer 1

Correct Answer: 935

Concept: For heights comparable to Earth's radius, acceleration due to gravity is not constant. Use conservation of mechanical energy: \[ \frac{1}{2}mv^2 = mgh \text{ (invalid for large h)} \] Correct approach: \[ \frac{1}{2}mv^2 = GMm\left(\frac{1}{R} - \frac{1}{R+h}\right) = mgR^2\left(\frac{1}{R} - \frac{1}{R+h}\right) \] where \(GM = gR^2\).

Step 1: Energy conservation equation. \[ \frac{1}{2}mv^2 = mgR^2\left(\frac{1}{R} - \frac{1}{R+h}\right) \] Cancel \(m\) and simplify: \[ \frac{v^2}{2} = gR^2\left(\frac{R+h - R}{R(R+h)}\right) = gR^2\left(\frac{h}{R(R+h)}\right) = \frac{gRh}{R+h} \]

Step 2: Solve for \(h\). \[ \frac{v^2}{2} = \frac{gRh}{R+h} \] \[ v^2(R+h) = 2gRh \] \[ v^2R + v^2h = 2gRh \] \[ v^2R = h(2gR - v^2) \] \[ h = \frac{v^2R}{2gR - v^2} \]

Step 3: Substitute values. \[ v = 4 \, km/s = 4000 \, m/s \] \[ R = 6400 \, km = 6.4 \times 10^6 \, m \] \[ g = 9.8 \, m/s^2 \] \[ h = \frac{(4000)^2 \times (6.4 \times 10^6)}{2 \times 9.8 \times (6.4 \times 10^6) - (4000)^2} \] \[ = \frac{16 \times 10^6 \times 6.4 \times 10^6}{2 \times 9.8 \times 6.4 \times 10^6 - 16 \times 10^6} \] \[ = \frac{102.4 \times 10^{12}}{(125.44 - 16) \times 10^6} \] \[ = \frac{102.4 \times 10^{12}}{109.44 \times 10^6} \] \[ = \frac{102.4}{109.44} \times 10^6 \, m \] \[ \approx 0.935 \times 10^6 \, m = 935 \, km \]