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$r_A=1.62 \times 10^8 km$

Giancoli 6th Edition, Chapter 5, Problem 57


Chapter 5, Problem 57 is solved.

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Transcript for this Giancoli solution

We can compare Icarus to the earth with Kepler’s laws. We can say the period of the asteroid’s orbit squared divided by the period of the earth’s orbit squared equals the radius of the asteroid’s orbit cubed divided by the radius of the earth’s orbit cubed. We’ll solve for ‘ra’ by multiplying both sides by the radius of the earth’s orbit cubed and then we’ll simplify and cube root both sides as well in order to solve for the radius of the asteroid’s orbit. So we have radius of Icarus’ orbit equals the cube root of earth's orbital radius cubed times the asteroid’s orbital period squared divided by the earth’s orbital period squared. Substituting numbers to get our answer: cube root of one hundred and forty nine point six times ten raised to power six kilometers times four hundred and ten days divided by three hundred and sixty five days squared and the answer to this will be one point six two times ten raised to power eight kilometers. A little tip: if you have trouble square or cube rooting on your calculating on your calculator, a cube rooting tip is to say you have exponent one third, that’s the same as cube rooting.