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$v=7.9 \times 10^3 m/s$

Giancoli 6th Edition, Chapter 5, Problem 47

(1:60)

Chapter 5, Problem 47 is solved.

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Transcript for this Giancoli solution

Mount Everest is really high but it’s not that high, what I mean is the centripetal acceleration of this satellite that’s going to be launched from the top of Mount Everest is essentially going to be the same as ‘g’ so that means we can say the centripetal acceleration ‘v2 over ‘r’ equals ‘g’ and we’ll solve this for ‘v’ and get our answer which is the square root of ‘r’ times ‘g’ after you multiply both sides by ‘r’ and take the square root of both sides. So ‘v’ equals the square root of six point three eight times ten raised to power six meters times nine point eight meters per second squared giving us our answer seven point nine times ten raised to power three meters per second.

Comments

Where did you get 6.38 x 10^6 since the radius of the Earth is 6.38 x 10^3?

Hi devon, the difference is in the units. The Earth's radius is $6.38 \times 10^6 \textrm{ m}$, or $6.38 \times 10^3 \textrm{ km}$.

Cheers,
Mr. Dychko