Question

Earth 's radius is about 6400 km ... if we put a satellite in orbit 6400 km above Earth's surface (constant distance, circular orbit), [ a) what is Earth gravity 's intensity at the satellite's location? ] [ b) what should the satellite's orbital speed be? ] c) what should the satellite's orbital (time) period be? 3583 s = 0.995 hours 7936 s = 2.204 hours 10134 s = 2.815 hours 1141 s = 0.317 hours 50 672 s = 14.08 hours 14 332 s = 3.98 hours

          Earth 's radius is about 6400 km ... if we put a satellite in orbit 6400 km above Earth's surface (constant distance, circular orbit), [ a) what is Earth gravity 's intensity at the satellite's location? ] [ b) what should the satellite's orbital speed be? ] c) what should the satellite's orbital (time) period be?
3583 s = 0.995 hours
7936 s = 2.204 hours
10134 s = 2.815 hours
1141 s = 0.317 hours
50 672 s = 14.08 hours
14 332 s = 3.98 hours
        
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Earth 's radius is about 6400 km ... if we put a satellite in orbit 6400 km above Earth's surface (constant distance, circular orbit), [ a) what is Earth gravity 's intensity at the satellite's location? ] [ b) what should the satellite's orbital speed be? ] c) what should the satellite's orbital (time) period be?
3583 s = 0.995 hours
7936 s = 2.204 hours
10134 s = 2.815 hours
1141 s = 0.317 hours
50 672 s = 14.08 hours
14 332 s = 3.98 hours

Added by Aaron P.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Earth 's radius is about 6400 km ... if we put a satellite in orbit 6400 km above Earth's surface (constant distance, circular orbit), [ a) what is Earth gravity 's intensity at the satellite's location? ] [ b) what should the satellite's orbital speed be? ] c) what should the satellite's orbital (time) period be? 3583 s = 0.995 hours 7936 s = 2.204 hours 10134 s = 2.815 hours 1141 s = 0.317 hours 50 672 s = 14.08 hours 14 332 s = 3.98 hours
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Transcript

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00:01 All right, so let's say we have a satellite that is at an orbit of twice earth's orbital distance.
00:09 We want to know what is the period of the satellite's orbit going to be, given that the mass of earth will take to be 5 .98 times 10 to the 24th kilograms.
00:21 And the radius of earth will take to be 6 ,400 kilometers.
00:27 Then kepler's third law says that like the period squared is going to be like this orbital distance cubed times four pi squared over g times m where m is the mass of earth and everything.
00:41 So t is going to be the square root of all this.
00:45 In fact, we can write it as like 2 pi times the square root of r cubed over gm...
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