Respuesta :
Answer:
6.922 × 10¹⁰ J.
Explanation:
The satellite is much smaller than the earth and can be treated as a point mass.
GPE of the Satellite
Consider the equation for the gravitational potential energy of a point mass at a certain distance away from the center of a spherical mass:
[tex]\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}[/tex],
where
- The gravitational constant [tex]G \approx 6.67\times 10^{-11}\;\textbf{m}^{3}\cdot \textbf{kg}^{-1}\cdot \text{s}^{-2}[/tex],
- [tex]M[/tex] the mass of the sphere (the earth in this case,)
- [tex]m[/tex] the mass of the satellite, and
- [tex]r[/tex] is the separation between the point mass and the center of the sphere.
As a side note, the GPE here is the same as the work that an external force will do when it brings the point mass from infinitely far away to the current point in space. Gravity pulls the object inwards; the external force acts against gravity in the opposite direction of travel. Energy will be released. The work of the external force will be negative. Hence, the value of GPE shall be either zero (at infinitely far away) or negative.
Convert all values to SI units:
- Mass of the earth: [tex]M = 5.972\times 10^{24}\;\textbf{kg}[/tex];
- Mass of the satellite: [tex]m = 2.000\times 10^{3}\;\textbf{kg}[/tex];
- Initial separation: [tex]r_{\text{Initial}} = 6400 \;\text{km} = 6.400\times 10^{6}\;\textbf{m}[/tex];
- Final separation: [tex]r_{\text{Final}} = 6400 \;\text{km} + 800\;\text{km} = 7200\;\text{km} = 7.200\times 10^{6}\;\textbf{m}[/tex]. Note the the radius of the orbit is the the distance between the satellite and the ground plus the radius of the planet.
Initial GPE of the satellite:
[tex]\displaystyle \begin{aligned}\text{GPE(Initial)} &= -\frac{G \cdot M \cdot m}{r_{\text{Initial}}}\\ &=-\frac{(6.67\times 10^{-11})\times (5.972\times 10^{24})\times (2.000\times 10^{3})}{6.400\times 10^{6}} \\ &= -1.245\times 10^{11}\;\text{J}\end{aligned}[/tex].
Final GPE of the satellite:
[tex]\displaystyle \begin{aligned}\text{GPE(Final)} &= -\frac{G \cdot M \cdot m}{r_{\text{Final}}}\\ &=-\frac{(6.67\times 10^{-11})\times (5.972\times 10^{24})\times (2.000\times 10^{3})}{7.200\times 10^{6}} \\ &= -1.106\times 10^{11}\;\text{J}\end{aligned}[/tex].
KE of the Satellite
Gravity is the only force that act on the satellite. The velocity [tex]v[/tex] of the satellite in this orbit depends on the size of the net force on the satellite, which is the same as gravity on the satellite. In other words,
[tex]\displaystyle \frac{m\cdot v^{2}}{r} = \Sigma F = W = \frac{G\cdot M\cdot m}{r^{2}}[/tex],
where again, [tex]r = 6400 + 800 = 7200\;\text{km} = 7.200\times 10^{6}\;\text{m}[/tex].
Rearranging gives:
[tex]\displaystyle v^{2} = \frac{G\cdot M}{r}[/tex],
[tex]\displaystyle \begin{aligned} v &= \sqrt{\frac{G\cdot M}{r}} \\ &= \sqrt{\frac{(6.67\times 10^{-11})\times (5.972\times 10^{24})}{7.200 \times 10^{6}}} \\ &= 7.438\times 10^{3}\;\text{m}\cdot \text{s}^{-1}\end{aligned}[/tex].
As a result,
[tex]\displaystyle \text{KE} = \frac{1}{2}\; m \cdot v^{2} = \frac{1}{2} \times (2.000\times 10^{3}) \times (7.438\times 10^{3}) = 5.532\times 10^{10} \;\text{J}[/tex].
Energy required
Assume no energy loss while lifting the satellite into the space. The energy required will be the same as the change in GPE plus the change in KE (Final minus Initial, keep any negative sign.)
[tex]\begin{aligned} \text{Energy Required} &= (\text{Final GPE} - \text{Initial GPE}) + (\text{Final KE} - \text{Initial KE})\\ &=((-1.106\times 10^{11}) - (-1.245\times 10^{11})) + ((5.532\times 10^{10} \;\text{J})) \\&=6.922 \times 10^{10}\;\text{J}\end{aligned}[/tex].
