A satellite moves in a circular orbit about our earth exactly four times a day. Find its orbit, radius , speed and weight.
2 answers
A 10.0 kg mass is 1.00m to the left of the origin, and a 20.0 kg mass is 3.00 m to the right of the origin, both on x-axis. Find the gravitational potential energy and the gravitational field on x-axis at a) x =-2.0m b.) the origin, c.) x = 4.00 (set Zero P.E @ infinity in terms of G.
The orbital period of any satellite derives from
T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds, 6(3600) = 21,600, r = the orbit radius in seconds and µ = the earth;s gravitational constant.
Therefore,
21600 = 2(3.14)sqrt(r^3/1.407974x10^21)
from which r = 55,002,293 ft. = 10,417 miles.
The velocity derives from V = sqrt(µ/r)
V = sqrt(1.407974x10^21/55,002,293) =
The orbital height and period are totally independant or the weight.
T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds, 6(3600) = 21,600, r = the orbit radius in seconds and µ = the earth;s gravitational constant.
Therefore,
21600 = 2(3.14)sqrt(r^3/1.407974x10^21)
from which r = 55,002,293 ft. = 10,417 miles.
The velocity derives from V = sqrt(µ/r)
V = sqrt(1.407974x10^21/55,002,293) =
The orbital height and period are totally independant or the weight.
1 answer