Question

A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2

Answer 1

8 × 106 m from the centre of the Earth

Velocity of the rocket, v = 5 km/s = 5 × 103 m/s

Mass of the Earth, me=6.0 ×1024 kg

Radius of the Earth, Re=6.4 ×106 m

Height reached by rocket mass, m = h

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

$=\frac{1}{2}mv^2+(\frac{-GM_em}{R_e})$

At highest point h,

v=0

and, potential energy=-$\frac{NM_em}{R_e+h}$

Total energy of the rocket $=0+(\frac{-GM_em}{R_e+h})=\frac{GM_em}{R_e+H}$

From the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface = Total energy at height h

$\frac{1}{2}mv^2+(\frac{-GM_em}{R_e})=-\frac{GM_em}{R_e+h}$

$\frac{1}{2}v^2=GM_e(\frac{1}{R_e}-\frac{1}{R_e+h)}$

$=GM_e(\frac{R_e+h-R_e}{R_e(R_e+h)})$

$\frac{1}{2}×v^2=\frac{GM_eh}{R_e(R_e+h)}$× $\frac{R_e}{R_e}$

$\frac{1}{2}×v^2=\frac{gh_eh}{R_e+h}$

Where g= $\frac{GM}{R_e^2}$ =9.8 m/s^2( Acceleration due to gravity on the Earth's surface)

∴ v2(Re+h)=2gReh

v2Re=h=(2gRe-v2 )

$h=\frac{R_ev^2}{2gR_e-v^2}$

$=\frac{6.4×10^6×(5×10^3)^2}{2×9.8×6.4×10^6-(5×10^3)^2}$

$h=\frac{6.4×25×10^{12}}{100.44×10^6}=1.6×10^6\,m$

Height achieved by the rocket with respect to the centre of the Earth

=Re+h

=6.4×106+1.6×106

=8.0×106 m