Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 5, Problem 90P

(a)

To determine

The satellite’s instantaneous velocity at point C.

(a)

Expert Solution
Check Mark

Answer to Problem 90P

The satellite’s instantaneous velocity at point C is 3.07kmin the y direction.

Explanation of Solution

Write an expression to calculate the satellite’s instantaneous velocity at point C.

v=2π(r+h)T (I)

Here, the satellite’s instantaneous velocity at point C is v, the radius of Earth is r, the altitude of satellite from Earth is h and T is the period.

Conclusion:

Substitute 6371km for r, 35800km for h, and 86400s for T in equation (I) to find v.

v=2π(6371km+35800km)86400s=2π(42171km)86400s=3.07km/s

Thus, the satellite’s instantaneous velocity at point C is 3.07kmin the y direction.

(b)

To determine

The satellite’s average velocity for one quarter of an orbit, starting at A and ending at B.

(b)

Expert Solution
Check Mark

Answer to Problem 90P

The satellite’s average velocity for one quarter of an orbit, starting at A and ending at B is  2.76km/sat 45° above the x axis.

Explanation of Solution

Write an expression to calculate the satellite’s average velocity for one quarter of an orbit, starting at A and ending at B.

|vav|=|Δr|Δt=(r+h)2(T/4)=(42)(r+h)T (II)

Here, the satellite’s average velocity for one quarter of an orbit, starting at A and ending at B is |vav|, the displacement is |Δr| and the time taken is Δt.

Conclusion:

Substitute 6371km for r, 35800km for h, and 86400s for T in equation (II) to find |vav|.

|vav|=(42)(6371km+35800km)86400s=(42)(42171km)86400s=2.76km/s

Thus, the satellite’s average velocity for one quarter of an orbit, starting at A and ending at B is  2.76km/sat 45° above the x axis.

(c)

To determine

The satellite’s average acceleration for one quarter of an orbit starting at A and ending at B.

(c)

Expert Solution
Check Mark

Answer to Problem 90P

The satellite’s average acceleration for one quarter of an orbit starting at A and ending at B is 0.201m/s2 at 45° below the x axis.

Explanation of Solution

Write an expression to calculate the satellite’s average acceleration for one quarter of an orbit starting at A and ending at B.

|aav|=|Δv|Δt=|vBvA|(T/4)=(v)2+(v)2(T/4)=4v2T (III)

Here, the satellite’s average acceleration for one quarter of an orbit starting at A and ending at B is |aav|, the change in velocity is |Δv|, and the time taken is Δt.

Conclusion:

Substitute 3.07m/s for v, and 86400s for T in equation (III) to find |aav|.

|aav|=4(3.07m/s)286400s=17.4m/s86400s=0.201m/s2

Thus, the satellite’s average acceleration for one quarter of an orbit starting at A and ending at B is 0.201m/s2 at 45° below the x axis.

(d)

To determine

The satellite’s instantaneous acceleration at point D.

(d)

Expert Solution
Check Mark

Answer to Problem 90P

The satellite’s instantaneous acceleration at point D and 0.224m/s2in the +ydirection.

Explanation of Solution

Write an expression to calculate the satellite’s instantaneous acceleration at point D.

a=GM(r+h)2 (IV)

Here, the satellite’s instantaneous acceleration at point D is a, the gravitational constant is G, the mass of Earth is M.

Conclusion:

Substitute 6.674×1011Nm2/kg2 for G, 5.974×1024kg for M, 6371km for r, and 35800km for h in equation (IV) to find a.

a=(6.674×1011Nm2/kg2)(5.974×1024kg)(6371km+35800km)2=(6.674×1011Nm2/kg2)(5.974×1024kg)(42171km)2=0.224m/s2

Thus, the satellite’s instantaneous acceleration at point D and 0.224m/s2in the +ydirection.

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Chapter 5 Solutions

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