Communications and weather satellites are frequently found at \"geostationary\"

Question

Communications and weather satellites are frequently found at "geostationary" orbits so that they can stay directly above a part of the Earth that the satellite services. There orbital radii "r" from the center of the Earth slowly decay, and thus their period decreases, and must be corrected with boosters to keep them in position.

Calculate the radius of the orbit in Earth radii (units of 6.37x10^6m) from the following decayed period 22.1 in hours. You need to invoke Newton's Universal Law of gravitation to solve for speed v and use the basic definition of speed. You will need Earth's mass and remember to convert from hours to seconds.

One Earth radius is 6.37x10^6m. Thus an object at a distance of 9.55x10^6m from the center of the earth is at 1.50 Earth radii.

Explanation / Answer

newton law

GMm2/r^2 =m2 v^2/r

GM/r = v^2.........................(1)

2pi*r = v*t

v=2*pi*r/t ................. (2)

from 1 and 2
GM/r =(2*pi*r/t )^2;

6.67

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