Understanding Our Universe
Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
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Chapter 9, Problem 38QAP

(a)

To determine

The mass of Electra.

(a)

Expert Solution
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Answer to Problem 38QAP

The mass of Electra is 8.17×1018kg .

Explanation of Solution

Write the expression for semi-major axis of moon orbit.

a=r+d        (I)

Here, r is the radius of asteroid, d is the distance between moon and asteroid and a is the average distance between the objects.

Write the expression for orbital period using Kepler’s third law.

T2=(4π2GM)a3

Here, G is gravitational constant, M is the mass of Electra and T is the orbital period.

Rearrange the above expression in term of M .

M=(4π2G)a3T2        (II)

Conclusion:

Calculate the radius of asteroid.

r=182km2=91km

Substitute 91km for r and 1325km for d in equation (I).

a=91km+1325km=1416km

Substitute 1416km for a, 5.25days for T and 6.67×1011m3/kgs2 for G in equation (II).

M=((4π26.67×1011m3/kgs2)((1416km)(103m1km))3((5.25days)(24h1day)(3600s1h))2)=8.17×1018kg

Thus, the mass of Electra is 8.17×1018kg .

(b)

To determine

The density of Electra.

(b)

Expert Solution
Check Mark

Answer to Problem 38QAP

The density of Electra is 2.6×103kg/m3 .

Explanation of Solution

Write the expression for volume of asteroid.

V=43πr3        (III)

Here, V is the volume of sphere.

Write the expression for density of asteroid.

ρ=MV        (IV)

Here, ρ is the density of asteroid.

Conclusion:

Substitute 91km for r in equation (III).

V=43π(91km)3=3.16×106km3

Substitute 3.16×106km3 for V and 8.17×1018kg for M in equation (IV).

ρ=8.17×1018kg3.16×106km3(103m1km)3=2.6×103kg/m3

Thus, the density of Electra is 2.6×103kg/m3 .

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