Class Management I Help HW10 (Unit 5 Gravitation) Begin Date: 10/4/2017 12:00:00
ID: 1778630 • Letter: C
Question
Class Management I Help HW10 (Unit 5 Gravitation) Begin Date: 10/4/2017 12:00:00 AM-Due Date: 10/15/2017 1159:00 PM End Date: 12/14/2017 12.00:00 AM Parent Satellite Average orbital radius r(km) Period Tey) (km 1.01×1019 3.35x 1024 3.35x 1024 (10%) Problem 9: A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite Eah 0.07481 1.000 11.86 0.00485 (1773.19x10 0.00972 355d 3.20x10 0.0196 (7.16 d) 3 .0457 (16.19 d 3.20x1021 3.84x 10 remains above the same point on Earth (provided it orbits in the Sun Eath 1496x 10 Jupiter7.783x10 4.22x 105 Europa 6.71x 105 Ganymede 1.07x10 Callisto 1.88x 10 equatorial plane in the same direction as Earth's rotation). upiter o 3.19×1021 Otheexpertta.com & Calculate the radius of such an orbit based on the data for the moon in the figure in km. Grade Sammary Potential Submissions 044 100% 45 6 Attempts remaining 10 cotanO asin acos atan acotan sinh0 tanh0 co . Degrees O Radians Submit Hint I give upl Feedback: deduction per feedback Hints: 0to deduction per hint. Himts remaining3Explanation / Answer
for a central body of mass M and satellite of mass m in a circular orbit of radius R
gravitational force = centripital force
GMm/R^2 = mv^2/R
GM/R = v^2
now, period of revolution around the body is given by T
T = 2*pi*R/v
v = 2*pi*R/T
hence
GM/R = 4*pi^2*R^2/T^2
R^3 = GM*T^2/4*pi^2
for earth moon system
R^3/T^2 = 1.01*10^19
so GM/4*pi^2 = 1.01*10^19 ( where M is mass of earth )
hence
for a satellite at radius r, time period T = 24 hrs = 24*60*60 seconds
R^3 = 1.01*10^19*(24*60*60)^2
R = 4224574.30207 km