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Get Full Access to University Physics - 13 Edition - Chapter 13 - Problem 86cp
Get Full Access to University Physics - 13 Edition - Chapter 13 - Problem 86cp

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# (a)When an object is in a circular orbit of radius r

ISBN: 9780321675460 31

## Solution for problem 86CP Chapter 13

University Physics | 13th Edition

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University Physics | 13th Edition

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Problem 86CP

(a)When an object is in a circular orbit of radius r around the earth (mass ), the period of the orbit is T, given by Eq. (13.12), and the orbital speed is given by Eq. (13.10). Show that when the object is moved into a circular orbit of slightly larger radius where its new period is and its new orbital speed is where and are all positive quantities and [Hint: Use the expression before they have a second chance. Find the numerical value of t and explain whether it would be worth the wait.

Step-by-Step Solution:

Solution 86CP Step 1 a) The time period of the object can be given as, 3 T = 2r2 Gm E Gm E The orbital speed of the obv =t, r 2(r+r) 2 The new time periodT + T = Gm E 3 Then, we can rewrite the t(r + r)2as, 3 3 3 (r + r) = r 2(1 + r) 2 r 3 By binomial expansion, we can wrrt2(1 + r) 2 = r2(1 + 3r) r 2 r Or, r2 1 + 3r = r + 3 r r ( 2 r) 2 Therefore, the equation will change as, 3 3 r r 2(r + 2 ) T + T = Gm E 3 (2r 2) 2(3 r r) That isT + T = + 2 Gm E Gm E 3 (2r 2) We can write, = T Gm E 3 r r Then, T + T = T + Gm E Rearranging, T + T = T + 3 r Gm E r Gm E Cancel out the terms T on both sides, and we can write,r = v Therefore, the change in time period of the object if we raise the orbit by a distance of r, 3 r T = v

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