Force Rquired to Put a Rocket Into Orbit

It takes a large sustained force to lift a rocket of mass nbsp;
\[200 tonnes=200 \times 10^3 kg\]
  into orbit  
\[200=200 \times 10^3\]
  above the Earth. A force must be applied to lift the rocket against the force of the Earths gravity, and to give it the speed necessary to keep it in orbit.
We can use the equation  
\[Work \: Done= Force \times Distance\]
  to find the average force. The work done will be equal to the total increase in energy of the rocket ie the increase in kinetic energy plus the increase in gravitational potential energy.. Assume the rocket starts on the Earths surface and take the Earth as a sphere with a a mass of  
\[5.98 \times 10^{24} kg\]
  and a radius of  
\[6370 km=6370 \times 10^3 m\]
, and the rocket The Earth has a mass of  
\[5.98 \times 10^{24} kg\]
The increase in gravitational potential energy is
\[\begin{equation} \begin{aligned} \Delta GPE &= -\frac{GM_{EARTH}m_{ROCKET}}{R_{ORBIT}}-(-\frac{GM_{EARTH}m_{ROCKET}}{R_{EARTH}}) \\ &=-\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times 200 \times 10^3}{6370 \times 10^3+200 \times 10^3} \\ &- (-\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times 200 \times 10^3}{6370 \times 10^3+200 \times 10^3})\\ &= 3.820 \times 10^{11} J \end{aligned} \end{equation}\]
When the rocket is on the Earths equator, it speed is the same as a point on the Earths equator:  
\[v=\frac{2 \pi R_{EARTH}}{T}=\frac{2 \pi \times 6370 \times 10^3}{24 \times 60 \times 60}=463.24 m/s\]
Hence the kinetic energy of the rocket on the Earths surface is  
\[\frac{1}{2} \times 200 \times 10^3 \times 463.24^2=2.1459 \times 10^{10} J\]

To find the speed of the rocket in orbit set the centripetal force equal to the gravitational force:

\[\frac{1}{2}m_{ROCKET}v^2=\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times 200 \times 10^3}{2 \times 6570 \times 10^3}=6.2616 \times 10^{12} J\]
The increase in kinetic energy is  
\[6.2616 \times 10^{12}-2.1459 \times 10^{10}=6.240 \times 10^{12}J\]
The total increase in energy is  
\[3.820 \times 10^{11}+6.240 \times 10^{12}=6.6221 \times 10^{12}J\]
This is equal to the average forces times distance, hence
\[Fd=6.6221 \times 10^{12} \rightarrow F=\frac{6.6221 \times 10^{12}}{100 \times 10^3}=6.6221 \times 10^7N\]

Add comment

Security code