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marshall27 [118]
3 years ago
6

A satellite is orbiting Earth in an approximately circular path. It completes one revolution each day (86,400 seconds). Its orbi

tal distance is 22,500 kilometers from the Earth's center. What is the satellite's average speed?
Physics
1 answer:
stira [4]3 years ago
6 0
@AL2006 had answered this before: Well, first of all, wherever you got this question from has done 
a really poor job of question-writing.  There are a few assorted 
blunders in the question, both major and minor ones:

-- 22,500 is the altitude of a geosynchronous orbit in miles, not km.

-- That figure of 22,500 miles is its altitude above the surface, 
   not its radius from the center of the Earth. 

-- The orbital period of a synchronous satellite has to match 
    the period of the Earth's rotation, and that's NOT 24 hours.  
    It's about 3 minutes 56 seconds less ... about 86,164 seconds. 

Here's my solution to the question, using some of the wreckage 
as it's given, and correcting some of it.  If you turn in these answers
as homework, they'll be marked wrong, and you'll need to explain 
where they came from.  If that happens, well, serves ya right for
turning in somebody else's answers for homework.


The satellite is traveling a circle. The circle's radius is 26,200 miles
(not kilometers) from the center of the Earth, so its circumference 
is (2 pi) x (26,200 miles) = about 164,619 miles.

    Average speed = (distance covered) / (time to cover the distance)

                             = (164,619 miles) / day
                                (264,929 km)

                             =      6,859 miles per hour
                                  (11,039 km) 

                           =          1.91 miles per second
                                      (3.07 km)
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1.69\cdot 10^{10}J

Explanation:

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E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J

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E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J

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\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J

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