Gravitation: An artificial satellite of 1, 200Kg orbiting the Earth completes ea
ID: 1500161 • Letter: G
Question
Gravitation: An artificial satellite of 1, 200Kg orbiting the Earth completes each orbit in 131 minutes. The radius of the Earth is 6.38 Times 10^6 m. The mass of the Earth is 5.98 Times 10^24 kg. G = 6.67 Times 10^-11N m^2 kg^-2 Find the altitude of the satellite, i.e. its distance to the Earth surface. Kepler's 3^rd Law: T^2 = (4 pi ^2/GM)r^3 What is the value of g at the location of this satellite? Equate centripetal force and gravity F_cen = m v^2/r' F_g = mg Find the orbital speed of the satellite in m/s. v = 2 pi r/T Find the gravitational potential energy of the satellite. V_G= -G mM/rExplanation / Answer
a) Here, (131 * 60)2 = (4 * 3.14 * 3.14 * r3)/(6.67 * 10-11 * 5.98 * 1024)
=> r = 8.549 * 106 m
=> altitude of satelite = 8.549 * 106 - 6.38 * 106
= 2.169 * 106 m
b) value of g = (6.67 * 10-11 * 5.98 * 1024)/(8.549 * 106)2
= 5.457 m/sec2
c) orbital speed of satellite = (2 * 3.14 * 8.549 * 106)/(131 * 60)
= 6830.5 m/sec
d) gravitational potential energy = (6.67 * 10-11 * 5.98 * 1024 * 1200)/(8.549 * 106)
= 5.6 * 1010 J
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.