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Problem 80 Easy Difficulty

Communication satellites are placed in a geosynchronous orbit, i.e. in a circular orbit such that they complete one full revolution about the earth in one sidereal day $(23.934 \mathrm{h})$, and thus appear stationary with respect to the ground. Determine $(a)$ the altitude of these satellites above the surface of the earth, $(b)$ the velocity with which they describe their orbit. Give the answers in both SI and U.S. customary units.

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Video Transcript

for the gravitational force between the earth on the satellite and it's circular orbit about art. We have that the force acting between the earth in the satellite is G capital M the mass of the Earth little M the mass of the satellite over there, separation off the centers R squared. And because of the circular orbit off the satellite, he forced you to the center of the centripetal force. Is MV squared over R. So from here, if you rearrange, we can get the speed of the satellite V to be the square root off G mm over our. But we also know that V Times Tour is equal to two pi r. So tall is the period off one orbit of the satellite and two pi r is the circumference or the total distance that this satellite recover. And if we square both sides, we get that V squared. Torre squared is equal to G and tall squared over our and this is equal to for pi squared r squared And so we can rearrange this equation and fine are so we get our cute We multiply both sides by our is equal to G am toss squared over for pi squared. Until then. From here we can see that the separation between the earth and the satellite is G I am Torre squared over four pi squared to the power off one third. So the cube root off gm tall, squared over four pi squared. Now that we have an expression for the separation, we can substitute our data into this equation. So Tor, the period off the satellite is 23 0.934 hours and this is equal to 86 point 16 24 times 10 to the three in seconds. So firstly, we're going to use S I units in S I units G is equal to nine for 81 m per square. Second, the radius of the earth are is equal to six 0.37 times 10 to the 6 m G times M, we know is equal to the gravitational acceleration little G and the square of the radius of the earth elsewhere. And this is equal to nine when 81 time, six 0.37 times 10 to the six squid. And this becomes 398 0.6 times 10 to the public 12 cubic meter per square second. So therefore, using the equation above, we confined our in S I units to be 398 0.0 six times 10 to the 12 That's needs a cube second squared or GM times 86 point 16 24 times 10 to the three squared over for hi squared and all of this to the power one third. And so if you perform this, we get are to be 42.145 times 10 to the 6 m. So the separation between the satellite and the sent off the earth It's 42.145 times 10 to the 6 m. What? We want to calculate the altitude. The altitude off the satellite above the earth will call it H is equal to this distance minus the radius off the earth since the altitude is the height above ground, not the height above the center off the earth. So this is if we take this value minus big are we get this to be 35 point 775 times 10 to the 6 m. So this is actually 35,000 800 kilometers for the altitude of the satellite is 35,800 kilometers and we can redo the calculation using US customer units where G is 32.2 ft, the square second and the ladies off the earth big our is 3960 miles which is 20,909 times 10 to the 6 ft. And if we do the calculation or we simply convert this answer from kilometers two miles, we get that this is 35,800 kilometers or 22,200 miles in altitude. So for pardon me, we wish to calculate the units and getting us students so V is equal to the square root off GM overall and again, we know these values. So that's the square root off 398 when 06 I can stand to the 12 in S I units over art which is funny to point 145 times 10 to the sixth in meters and so we get the speed of the satellite to be 3.7 times 10 to the three meters per second and this is simply 3.7 kilometers per second. So we can do the calculation again and we get reconvert. These are units into US units and we get the velocity to be 10 0.9 times 10 to the power three feet the second.

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Vector Mechanics for Engineers: Statics and Dynamics

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Video Transcript

Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

Andy Chen

Liev Birman

Marshall Styczinski

Zachary Mitchell

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Vector Mechanics for Engineers: Statics and Dynamics

Andy Chen

Liev Birman

Marshall Styczinski

Zachary Mitchell

Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek Vector Mechanics for Engineers: Statics and Dynamics

Andy Chen

Liev Birman

Marshall Styczinski

Zachary Mitchell

Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek

Vector Mechanics for Engineers: Statics and Dynamics