Question

A satellite of mass mS orbits the Earth (with mass mE and radius RE) with a velocity v and an

altitude h. The gravitational force FG and the centripetal force FC are given by:

FG = G

mS · mE

(RE + h)

2

, FC =

mS · v

2

RE + h

, G = const.

(a) Find an equation for the kinetic energy Ekin(h) of a satellite with an altitude h.

(b) Based on the kinetic energy, how much liquid hydrogen (energy density: 106

J/Litre) is at least

needed to bring a small 1 kg satellite in an orbit of 400 km. (Use literature to find mE, RE, G.)​

Answer 1

Answer:

Part a)

Kinetic energy of satellite is given as

$KE = \frac{GM_eM_s}{2(R_e + h)}$

Part b)

Total volume of the fuel required is

$V = 59 Ltr$

Explanation:

As we know that the gravitational force on the satellite is providing the required centripetal force on it

So we have

$F = \frac{GM_e M_s}{(R_e + h)^2}$

here we know that this is centripetal force on it

so we have

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

so we have

$KE = \frac{1}{2}M_sv^2$

$KE = \frac{GM_eM_s}{2(R_e + h)}$

Part b)

As we know that total kinetic energy is given as

$KE = \frac{GM_eM_s}{2(R_e + h)}$

now plug in all values in it

$KE = \frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})(1)}{(6.37 \times 10^6 + 4\times 10^5)}$

$KE = 5.9 \times 10^7 J$

now we know that energy density of the fuel is

$u = 10^6 J/L$

so total volume required

$V = \frac{KE}{u}$

$V = 59 Ltr$

#Learn

Topic : gravitational potential energy

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