Answer :
Answer:
Approximately [tex]5.4 \times 10^{8}\; {\rm J}[/tex] of energy will be required. Total mechanical energy of this satellite is equal to [tex](1/2)[/tex] of its gravitational potential energy.
The kinetic energy of the system will decrease by approximately [tex]5.4 \times 10^{8}\; {\rm J}[/tex] (a change of [tex](-5.4) \times 10^{8}\; {\rm J}[/tex].)
The gravitational potential energy of the system will increase by approximately [tex]1.1\times 10^{9}\; {\rm J}[/tex].
Explanation:
Let [tex]G[/tex] denote the gravitational constant. Let [tex]M[/tex] denote the mass of the Earth. Let [tex]r[/tex] denote the radius of the orbit. Let [tex]v[/tex] denote the orbital speed of the satellite. Let [tex]m[/tex] denote the mass of the satellite.
Assume that the gravitational attraction from the Earth [tex](G\, M\, m) / (r^{2})[/tex] is the only force on the satellite. The acceleration of the satellite will be:
[tex]\begin{aligned}(\text{acceleration}) &= \frac{(\text{net force})}{(\text{mass})} \\ &= \frac{(G\, M\, m) / (r^{2})}{m}= \frac{G\, M}{r^{2}}\end{aligned}[/tex].
Since the satellite is in a uniform circular motion of orbital speed [tex]v[/tex] and radius [tex]r[/tex], acceleration will be [tex](v^{2} / r)[/tex] (centripetal acceleration.)
[tex]\begin{aligned}(\text{acceleration}) &= \frac{v^{2}}{r}\end{aligned}[/tex].
Therefore:
[tex]\begin{aligned}\frac{v^{2}}{r} &= (\text{acceleration}) = \frac{G\, M}{r^{2}}\end{aligned}[/tex].
Rearrange and solve for orbital speed [tex]v[/tex]:
[tex]\begin{aligned}v^{2} &= \frac{G\, M}{r}\end{aligned}[/tex].
[tex]\begin{aligned}v &= \sqrt{\frac{G\, M}{r}}\end{aligned}[/tex].
With a mass of [tex]m[/tex], the kinetic energy [tex](\text{KE})[/tex] of this satellite will be:
[tex]\begin{aligned}(\text{KE}) &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2}\, m\, \left(\frac{G\, M}{r}\right) = \frac{G\, M\, m}{2\, r}\end{aligned}[/tex].
The gravitational potential energy [tex](\text{GPE})[/tex] of the satellite would be:
[tex]\begin{aligned}(\text{GPE}) &= \left(-\frac{G\, M\, m}{r}\right)\end{aligned}[/tex].
The total mechanical energy of this satellite is the sum of [tex](\text{KE})[/tex] and [tex](\text{GPE})[/tex].
[tex]\begin{aligned}(\text{KE}) + (\text{GPE}) &= \left(\frac{G\, M\, m}{2\, r}\right) + \left(-\frac{G\, M\, m}{r}\right) = \frac{-G\, M\, m}{2\, r}\end{aligned}[/tex].
Hence, the total (mechanical) energy of this satellite is [tex](1/2)[/tex] the value of potential energy.
The radius of the Earth is approximately [tex]6.378 \times 10^{6}\; {\rm m}[/tex]. At the initial orbit ([tex]h = 93\; {\rm km} = 93\times 10^{3}\; {\rm m}[/tex],) the orbital radius will be approximately [tex]6.378 \times 10^{6}\; {\rm m} + 9.3 \times 10^{4}\; {\rm m} \approx 6.471 \times 10^{6}\; {\rm m}[/tex].
- Kinetic energy:
[tex]\begin{aligned} & (\text{KE}) \\ =\; & \frac{G\, M\, m}{2\, r} \\ =\;& (6.6743 \times 10^{-11}\; {\rm m^{3}\cdot kg^{-1} \cdot s^{-2}})\\ & (5.792\times 10^{24}\; {\rm kg}) \\ & (1020\; {\rm kg}) \\ & (1/2) \\ & (1/(6.471 \times 10^{6}\; {\rm m})) \\ \approx\; & 3.0467\times 10^{12}\; {\rm J}\end{aligned}[/tex]. - Gravitational potential energy:
[tex]\begin{aligned} & (\text{GPE}) = \frac{-G\, M\, m}{r} \approx (-6.0934) \times 10^{10}\; {\rm J}\end{aligned}[/tex]. - Total mechanical energy:
[tex]\begin{aligned} (\text{KE}) + (\text{GPE}) =\; & \frac{-G\, M\, m}{2\, r} \approx (-3.0467) \times 10^{10}\; {\rm J}\end{aligned}[/tex].
At the new orbit ([tex]r = 210\; {\rm km} = 210 \times 10^{3}\; {\rm m}[/tex]):
- Kinetic energy:
[tex]\begin{aligned} (\text{KE}) = \frac{G\, M\, m}{2\, r} \approx 2.9926 \times 10^{10}\; {\rm J}\end{aligned}[/tex]. - Gravitational potential energy:
[tex]\begin{aligned} (\text{GPE}) =& \frac{-G\, M\, m}{r} \approx(-5.9852\times 10^{10})\; {\rm J}\end{aligned}[/tex]. - Total mechanical energy:
[tex]\begin{aligned} (\text{KE}) + (\text{GPE}) = -\frac{G\, M\, m}{2\, r} \approx\; &(-2.9926) \times 10^{10}\; {\rm J}\end{aligned}[/tex].
The energy that need to be added to the system is equal to the difference in total mechanical energy ([tex](\text{GPE}) + (\text{KE})[/tex]):
[tex]((-2.9926) \times 10^{10}\; {\rm J}) - ((-3.0467) \times 10^{10}\; {\rm J}) \approx 5.4 \times 10^{8}\; {\rm J}[/tex].
Change in the kinetic energy of the system:
[tex](2.9926 \times 10^{10}\; {\rm J}) - (3.0467\times 10^{10}\; {\rm J}) \approx (-5.4) \times 10^{8}\; {\rm J}[/tex].
In other words, the kinetic energy of the system would be reduced by approximately [tex](-5.4) \times 10^{8}\; {\rm J}[/tex].
Change in the gravitational potential energy of the system:
[tex]\begin{aligned}& (-5.9852\times 10^{10}\; {\rm J}) - (-6.0934\times 10^{10}\; {\rm J})\approx 1.1 \times 10^{9}\; {\rm J}\end{aligned}[/tex].
In other words, the gravitational potential energy of the system would increase by approximately [tex]1.1 \times 10^{9}\; {\rm J}[/tex].
