recategorized Mar 6, 2020 by subrita Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [g is acceleration due to gravity at Earth’s surface] (a) 2π √2R/g (b) 4√2π√R/g Paiye sabhi sawalon ka Video solution sirf photo khinch kar. The equation assumes that the satellite is high enough off the ground that it orbits out of the atmosphere. http://en.wikipedia.org/wiki/Geostationary_orbit, http://en.wikipedia.org/wiki/Sidereal_day. Note that the distance of the satellite from the Earth (r) in the formula is the distance from the center. This latitude will never exceed I in absolute magnitude, and is determined by the anomaly θ according to Enter 12 h for P and 1 M_Earth for M and 1 R_Earth for r . A satellite is revolving in a circular orbit at a height h from the Earth's surface (radius of Earth R,h< Knights Estate Agents Barry, La Rams Front Office, Costa Coffee Club Problems, Plymouth State University Calendar 2020-2021, City Of St Charles, Il Brush Pickup Schedule, Authentic Jade Roller, Bedworth Council Phone Number, Plymouth State University Calendar 2020-2021, Mahalo Ukulele - Pink,