Respuesta :
Answer:
Explanation:
Time period of satellite
a )T = [tex]\frac{2\pi\times(R+h)^\frac{3}{2} }{\sqrt{GM}}[/tex]
Where G is gravitational constant , M is mass of the earth , R is radius of the earth and h is height of satellite .
[tex]\frac{2\pi\times(6378+850)^\frac{3}{2}\times10^{4.5} }{\sqrt{6.67\times10^{-11}\times5.97\times10^{24}} }[/tex]
= 6273 s
= 104.5 minutes
b )
Orbital speed of satellite
v_o = [tex]\sqrt{\frac{GM}{(R+h)} }[/tex]
= [tex]\sqrt{\frac{6.6\times10^{-11}\times5.97\times10^{24}}{(6378+850)\times10^3} }[/tex]
= 7.38 km /s
c )
Orbital energy
= [tex]\frac{-GM}{2(R+h)} }[/tex]
= [tex]\frac{-6.67\times10^{-11}\times5.67\times10^{24}}{2(6378+850)\times10^3} }[/tex]
= - 2.6 x 10⁷ J
Potential energy on the surface
= - GM / R
=- 6.67 x 10⁻¹¹ x 5.67 x 10²⁴ / 6378 x 10³
- 5.93 x 10⁷
Energy required to place the satellite in orbit at height h
= 5.93 x 10⁷ - 2.6 x 10⁷
= 3.33 x 10⁷ J
Assuming a circular orbit, the time period required by the satellite to complete one orbit is 6118.59 secs.
Given the following data:
- Mass of satellite = 190 kg.
- Height = 850 km.
Scientific data:
- Radius of Earth = [tex]6.37\times 10^6\;m[/tex].
- Gravitational constant = [tex]6.67 \times 10^{-11}[/tex]
- Mass of Earth = [tex]5.972\times 10^{24}[/tex] kg
How to calculate the time period.
From Kepler’s Law for elliptic motion, the time period required by the satellite to complete one orbit is given by this formula:
[tex]T=\sqrt{\frac{4\pi^2 R^3}{GM} } \\\\T=\sqrt{\frac{4\pi^2 (850 \times 10^3+6.37\times 10^6)^3}{6.67 \times 10^{-11} \times 5.972\times 10^{24}} }\\\\T=\sqrt{\frac{1.49 \times 10^{22} }{3.98 \times 10^{14} } } \\\\T=\sqrt{37437185.93 }[/tex]
T = 6118.59 secs.
b. To determine the satellite's speed:
Mathematically, the satellite's speed is given by this formula:
[tex]V=\frac{2\pi \times (850 \times 10^3+6.37\times 10^6)}{6118.59} \\\\V=\frac{45370480}{6118.59}[/tex]
V = 7415 m/s.
c. To determine the minimum energy input that is necessary to place this satellite in orbit:
[tex]\Delta E=\frac{1}{2} m(v^2-v_e^2)+GMm(\frac{1}{r} -\frac{1}{R} )\\\\\Delta E=\frac{1}{2} \times 190(7415^2-463.24^2)+6.67 \times 10^{-11} \times 5.972\times 10^{24} \times 190 (\frac{1}{6.37 \times 10^6} -\frac{1}{850 \times 10^3+6.37\times 10^6} )\\\\\Delta E=95(54767633.7)+7.57\times 10^{16}(1.57\times 10^{-7}-1.39\times 10^{-7})\\\\\Delta E=5202925201.5+7.57\times 10^{16}(1.8\times 10^{-8})\\\\\Delta E=5202925201.5+1362600000\\\\\Delta E=6.57 \times 10^{9}\;Joules.[/tex]
Read more on Kepler's law here: brainly.com/question/6867220
