Question

based on the kinetic energy, how much liquid hydrogen (energy density: 10^6 J/Litre) is at least needed to bring a small 1kg satellite in an orbit of 400km. use literature to find mass, radius)

Answer 1

Answer:

Volume of the fuel that is required is 33 Ltr

Explanation:

Initial total energy of the satellite when it is on the surface of Earth

$E = K + U$

$E = 0 -\frac{GMm}{R}$

Final total energy when it is placed in the orbit is given as

$E_f = \frac{GMm}{2(R + h)} - \frac{GMm}{R + h}$

so we have

$E_f = -\frac{GMm}{2(R + h)}$

now we have

$\Delta E = E_f - E_i$

$\Delta E = GMm(\frac{1}{R} - \frac{1}{2(R+h)}$

$\Delta E = (6.67\times 10^{-11})(5.98 \times 10^{24})(1)(\frac{1}{6.4 \times 10^6} - \frac{1}{2(6.8\times 10^6)})$

$\Delta E = 3.3 \times 10^7 J$

now we know that the energy density is given as

$u = 10^6 J/Ltr$

so total mass required is

$V = \frac{3.3 \times 10^7}{10^6} Ltr$

$V = 33 Ltr$

#Learn

Topic : Energy of satellite

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