- #1

Nirmal Padwal

- 41

- 2

- Homework Statement
- Expressing kinetic energy of a satellite in a circular orbit in terms of the height of the satellite above the earth's surface (##h##), calculate the amount of hydrogen (energy density: ##10^6## J/Litre) needed to bring a small 1.5 kg satellite in an orbit of 200 km. (Given: ##m_e## = 5.972##\times 10^{24}## kg, ##R_e## = 6371 km = 6371##\times 10^3## m, ##G## = 6.67##\times 10^{-11}## N.m##^2##.kg##^{-2}##

- Relevant Equations
- 1. $$K.E. = G \frac{m_s . m_e}{2(R_e+h)}\\$$

2. $$ \mathrm{Required \ amount} = \frac{E_{kin}(h)}{E_{dens}}$$

The kinetic energy of a ##m_s##= 1.5 kg satellite in an orbit of ##h## = 200 km = 200##\times 10^3## m is

\begin{eqnarray*}

K.E. &=& G \frac{m_s . m_e}{2(R_e+h)}\\

&=& (6.67 \times 10^{-11} \ \mathrm{N.m^2.kg^{-2}}) \times

\frac{(1.5\ \mathrm{kg})\times (5.972\times 10^{24}\ \mathrm{kg})}{2\times (6371+200) \times 10^3

\ \mathrm{m}}\\

E_{kin}(h) &=& 45.465 \times 10^6\ \mathrm{J}

\end{eqnarray*}

As energy density of liquid hydrogen is given to be ##E_{dens}## = ##10^6## J/Litre, total amount of liquid hydrogen required for the satellite to gain kinetic energy of 45.465 ##\times 10^6## J is

$$ \mathrm{Required \ amount} = \frac{E_{kin}(h)}{E_{dens}} = \frac{45.465 \times 10^6\ \mathrm{J}}{10^6 \ \mathrm{J/Litre}}= 45.465 \ \mathrm{Litre}$$

I am not sure if this solution is right. Please let me know if something is not right.

\begin{eqnarray*}

K.E. &=& G \frac{m_s . m_e}{2(R_e+h)}\\

&=& (6.67 \times 10^{-11} \ \mathrm{N.m^2.kg^{-2}}) \times

\frac{(1.5\ \mathrm{kg})\times (5.972\times 10^{24}\ \mathrm{kg})}{2\times (6371+200) \times 10^3

\ \mathrm{m}}\\

E_{kin}(h) &=& 45.465 \times 10^6\ \mathrm{J}

\end{eqnarray*}

As energy density of liquid hydrogen is given to be ##E_{dens}## = ##10^6## J/Litre, total amount of liquid hydrogen required for the satellite to gain kinetic energy of 45.465 ##\times 10^6## J is

$$ \mathrm{Required \ amount} = \frac{E_{kin}(h)}{E_{dens}} = \frac{45.465 \times 10^6\ \mathrm{J}}{10^6 \ \mathrm{J/Litre}}= 45.465 \ \mathrm{Litre}$$

I am not sure if this solution is right. Please let me know if something is not right.