Respuesta :
GIVEN DATA
Mass of Rocket = m = 215 kg
Mass of Earth = M = [tex]5.97*10^{24}[/tex] kg
Radius of Earth = R = [tex]6.37*10^6[/tex] m
Gravitational Constant = G = [tex]6.67*10^-11 \;\;m^3kg^{-1}s^{-2}[/tex]
Speed of Rocket = 6.42 km/s
Initial Height of Rocket from Earth's Surface = 307 km = [tex]3.07*10^5[/tex] m
Final Height of Rocket from Earth's Surface = 731 km = [tex]7.31*10^5[/tex] m
Initial Height from Earth's Centre = [tex]R_i[/tex] = [tex]3.07*10^5 + 6.37*10^6 = 6.677*10^6[/tex] m
Final Height from Earth's Surface = [tex]R_f[/tex] = [tex]7.31*10^5 + 6.37*10^6 = 7.101*10^6[/tex] m
(a) Kinetic Energy at Final Height = [tex]K.E_f[/tex]
(b) Maximum Height of Rocket above Earth's Surface = [tex]H_{max}[/tex]
EXPLANATION
Part (a):
As drag air is negligible, energy will be conserved.
[tex]\therefore[/tex] ΔE = 0
[tex]K.E_i + U_i = K.E_f + U_f[/tex]
[tex]K.E_f = U_f - K.E_i - U_i[/tex]
where, U is the potential energy of the system.
[tex]K.E_f=\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}\;\;\;\;----------\;(1)[/tex]
Substituting Values and simplifying,
[tex]K.E_f = 3.665*10^9 J[/tex]
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.
[tex]i.e.\;\;v_f = 0\;\;\;\&\;\;\; K.E_f=0[/tex]
Equation (1) will become,
[tex]0\;=\;\;G\frac{mM}{R_f} + \frac{1}{2}mv_i^2\;-\;G\frac{mM}{R_i}[/tex]
[tex]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}[/tex]
Substituting values and simplifying,
[tex]R_f = 10.20*10^6 m[/tex]
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.
[tex]H_{max} = R_f - R[/tex]
[tex]H_{max} = 10.20*10^6\;-\;6.37*10^6\\\\H_{max} = 3.83*10^6\;\;m[/tex]
