I have my doubts about your signs etc.
Here:
https://www.youtube.com/watch?v=iaCw6A2s96w
also check the replies below
Many satellites orbit the earth at about 1000km above the earth's surface. Geosynchronous satellites orbit a distance of 4.22*10^7 m from the center of the earth. How much more energy is required to launch a 500kg satellite into a geosynchronous orbit rather than launching into an orbit 1000km above the earth's surface.
mass of earth = 5.972*10^24 kg
mass of satellite = 500 kg
radius of earth = 6371 km = 6.371*10^6 m
radius of satellite orbit = 6371km + 1000km = 7371km = 7.371*10^6 m
radius of geo satellite orbit = 4.22*10^7 m
mu = (m_earth)(m_satellite)/(m_earth + m_satellite) = 500kg = m_satellite
I am assuming that the orbit of the satellite around the earth is circular.
For the satellite orbit:
E_total = KE + Ug
E_total = (L^2)/(2)(mu)(r_satellite^2) - (G)(m_earth)(m_satellite)/(r_satellite)
= (L^2)/(2)(mu)(r_satellite^2) - (G)(m_earth)(m_satellite)(2)(mu)(r_satellite)/(2)(mu)(r_satellite^2)
= (L^2)/(2)(mu)(r_satellite^2) - (G)(m_earth)(m_satellite^2)(2)(r_satellite)/(2)(mu)(r_satellite^2)
Using G = 6.67*10^-11 Nm^2/kg^2 and known values mentioned above,
E_total = (L^2 - 1.46805406*10^27)/(5.4331641*10^16)
For the geosynchronous satellite orbit:
I used the same equation here as I did for the satellite orbit, replacing the r_satellite with r_geo satellite to get:
E_total = (L^2 - 8.40481364*10^27)/(1.78084*10^18)
Let L = 5.00*10^10 kgm^2/s:
E_total satellite orbit = -2.702019547*10^10 J
E_total geo satellite orbit = -4.719576795*10^9 J
Then I would compare the difference of those two final values, but I'm not even sure if I'm doing this correctly at all. Another option I tried was determining the values of the gravitational potential energy (Ug) of each orbit, which would result in negative values in both cases. The energy required would be the positive equivalent values and then I would find the difference. Both options give very similar values, only differing towards the far right decimal places.
2 answers
By the way I assume you launch for a location on the equator where even a statue is moving about a thousand miles per hour toward the East, although that initial spin KE does not matter much i the overall scheme of things.
1 answer