Question

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.​

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

Answer 2

Answer:

okh....

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