Answer in Physics for sabby #254179
January 23rd, 2023
What is the speed of a satellite moving in a circular orbit at a height of 3800km above
the surface of the earth? (b) what is the period of the satellite in hours? The mass of the
earth is 5.97×10^24 kg. The radius of the earth is 6.38×10^6 m.
Given:
“G=6.67*10^{-11}:rm Ncdot m^2/kg^2”
“R_E=6.38*10^{6}:rm m”
“h=3.80*10^{6}:rm m”
“M_E=5.97*10^{24}:rm kg”
(a) the speed of a satellite moving in a circular orbit
“v=sqrt{frac{GM_E}{R_E+h}}”
“v=sqrt{frac{6.67*10^{-11}*5.97*10^{24}}{6.38*10^{6}+3.80*10^{6}}}=6250:rm m/s”
(b) what is the period of the satellite in hours
“T=frac{2pi(R_E+h)}{v}”
“T=frac{2pi(6.38*10^{6}+3.80*10^{6})}{6250}=10200:rm s=2.84: hr”
