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(Rp +h)2 Fc = mg. v2G=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:

Kinetic energy of satellite is given as

$KE = \frac{GM_e m}{2(R + h)}$

Amount of fuel required is 29.5 ltr

Explanation:

As we know that the gravitational force on the satellite is the centripetal force on it to revolve in circle

So here we have

$\frac{GM_em}{(R + h)^2} = \frac{mv^2}{(R + h)}$

so here we have

$mv^2 = \frac{GM_em}{(R + h)}$

now the kinetic energy of the satellite is given as

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

$KE = \frac{GM_e m}{2(R + h)}$

As we know that the KE of the satellite is given as

$KE = \frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})(1)}{2(6.36 \times 10^6 + 4 \times 10^5)}$

so we have

$KE = 2.95 \times 10^7 J$

Now we know that energy density is given as

$u = 10^6 J/L$

now total amount of fuel required is given as

$V = \frac{2.95 \times 10^7}{10^6}$

$V = 29.5 L$

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Topic : Energy of Satellite

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