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, m_e=6.0 ×10²⁴  kg

Radius of the Earth,  R_e=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(R_e+h)=2gR_eh

v²R_e=h=(2gR_e-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

=R_e+h

=6.4×10⁶+1.6×10⁶

=8.0×10⁶ m