Question

A 215 kg rocket moving radially outward from Earth has a speed of 6.42 km/s when its engine shuts off 307 km above Earth's surface. (a) Assuming negligible air drag, find the rocket's kinetic energy when the rocket is 731 km above Earth's surface. (b) What maximum height above the surface is reached by the rocket

Answer 1

GIVEN DATA

Mass of Rocket = m = 215 kg

Mass of Earth = M = $5.97*10^{24}$ kg

Radius of Earth = R = $6.37*10^6$ m

Gravitational Constant = G = $6.67*10^-11 \;\;m^3kg^{-1}s^{-2}$

Speed of Rocket = 6.42 km/s

Initial Height of Rocket from Earth's Surface = 307 km = $3.07*10^5$ m

Final Height of Rocket from Earth's Surface = 731 km = $7.31*10^5$ m

Initial Height from Earth's Centre = $R_i$ = $3.07*10^5 + 6.37*10^6 = 6.677*10^6$ m

Final Height from Earth's Surface = $R_f$ = $7.31*10^5 + 6.37*10^6 = 7.101*10^6$ m

(a) Kinetic Energy at Final Height = $K.E_f$

(b) Maximum Height of Rocket above Earth's Surface = $H_{max}$

EXPLANATION

Part (a):

As drag air is negligible, energy will be conserved.

$\therefore$ ΔE = 0

$K.E_i + U_i = K.E_f + U_f$

$K.E_f = U_f - K.E_i - U_i$

where, U is the potential energy of the system.

$K.E_f=\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}\;\;\;\;----------\;(1)$

Substituting Values and simplifying,

$K.E_f = 3.665*10^9 J$

Part (b):

The rocket will come to rest after reaching the maximum height. Therefore, its final velocity and consequently final kinetic energy will be zero.

$i.e.\;\;v_f = 0\;\;\;\&\;\;\; K.E_f=0$

Equation (1) will become,

$0\;=\;\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}$

$or\;\;\frac{1}{2}v_i^2= GM(\frac{1}{R_i}-\frac{1}{R_f})\\\\\frac{1}{2GM}v_i^2= (\frac{1}{R_i}-\frac{1}{R_f})\\\\\therefore\; R_f = \frac{2GMR_i}{2GM-v_i^2R_i}$

Substituting values and simplifying,

$R_f = 10.20*10^6 m$

which is the distance from the Earth's centre. To find the height of rocket from Earth's surface, we simply subtract the Earth's radius from above result.

$H_{max} = R_f - R$

$H_{max} = 10.20*10^6\;-\;6.37*10^6\\\\H_{max} = 3.83*10^6\;\;m$