Answer:
1.3*10^14 J
Explanation:
The energy of the satellite that orbits the earth is given by the second Newton law:
[tex]F=ma_c\\\\-G\frac{mM_s}{r^2}=m\frac{v^2}{r}\\\\v^2=\frac{GM}{r}\\\\E_T=K+U=G\frac{mM_s}{2r}-G\frac{mM}{r}=-G\frac{mM}{2r}[/tex]
where you have taken into account the centripetal acceleration of the satellite.
m: mass of the satellite
M_s: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.67*10^-11 m^3/kg s^2
r: distance to the center of the Earth = Earth radius + distance satellite-Earth surface
To find the needed energy, you first compute the energy for a constant altitude of 99km:
r = 6.371*10^6m + 99*10^3m = 6.47*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.47*10^6)}\\\\E_T=-9.88*10^{15} \ J[/tex]
Next, you calculate the energy for an altitude of 195km:
r = 6.371*10^6m + 195*10^{3}m = 6.56*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.56*10^6)}\\\\E_T=-9.75*10^{15} \ J[/tex]
Finally, the energy required to put the satellite in the new orbit is:
-9.75*10^15 J - (-9.88*10^15 J) = 1.3*10^14 J
