Given:
The orbital height of the satellite, h=94 km=94000 m
The mass of the satellite, m=1045 kg
The new altitude of the satellite, d=207 km=207000 m
To find:
a) The energy needed.
b) The change in the kinetic energy.
c) The change in the potential energy.
Explanation:
The radius of the earth, R=6.37×10⁶ m
The mass of the earth, M=6×10²⁴ kg
a) The orbital velocity is given by,
[tex]v=\sqrt{\frac{GM}{r}}[/tex]Where G is the gravitational constant and r is the radius of the satellite from the center of the earth.
Thus the initial orbital velocity of the earth,
[tex]\begin{gathered} v_1=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)}} \\ =7868.43\text{ m/s} \end{gathered}[/tex]The orbital velocity after changing the altitude is,
[tex]\begin{gathered} v_2=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)}} \\ =7800.5\text{ m/s} \end{gathered}[/tex]Thus the total energy needed is given by,
[tex]E=(\frac{1}{2}mv_2^2-\frac{GMm}{(R+d)})-(\frac{1}{2}mv_1^2-\frac{GMm}{(R+h)})[/tex]On substituting the known values,
[tex]\begin{gathered} E=1045[(\frac{1}{2}\times7868.43^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)})-(\frac{1}{2}\times7800.5^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)})] \\ =623\text{ MJ} \end{gathered}[/tex]b)
The change in the kinetic energy is given by,
[tex]\begin{gathered} KE=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 \\ =\frac{1}{2}m(v_2^2-v_1^2) \end{gathered}[/tex]On substituting the known values,
