To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:
Where,
G = Gravitational Universal Constant
M = Mass of Planet
r = Radius of the planet ('h' would be the orbit from the surface)
The escape velocity is
Through this equation we can find the mass of the Planet in function of the distance, therefore
The orbital velocity is
The time period of revolution is,
Therefore the orbital period of the satellite is closes to 1 hour and 12 min
Answer:
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Explanation:
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v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ;
and solve for "v".
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Explanation:
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The formula is: KE = (½) * (m) * (v²) ;
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"Kinetic energy" = (½) * (mass) * (velocity , "squared")
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Note: Velocity is similar to speed, in that velocity means "speed and direction"; however, if you "square" a negative number, you will get a "positive"; since: a "negative" multiplied by a "negative" equals a "positive".
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So, we have the formula:
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KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
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we arrange the formula ;
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(KE) = (½) * (m) * (v²) ; ↔ (½)*(m)* (v²) = (KE) ;
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→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
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→ We divide each side of the equation by: "[(½)* (m)]" ;
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→ [ (½)*(m)*(v²) ] / [(½)* (m)] = (KE) / [(½)* (m)]<span> ;
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to get:
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→ v² = (KE) / [(½)* (m)]
→ v² = 2 KE / m
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Take the "square root" of each side of the equation ;
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→ √ (v²) = √ { 2*(KE) ] / m }
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→ v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"];
and solve for "v".
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