21st Century Astronomy
21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 12, Problem 38QP

(a)

To determine

The mass of the asteroid Electra.

(a)

Expert Solution
Check Mark

Answer to Problem 38QP

The mass of the asteroid Electra is 1.269×1019kg .

Explanation of Solution

Write the expression for the Kepler’s law for the time period of a celestial body around the planet.

  T2=(4π2GM)R3

Rearrange the above expression for the mass of the planet.

  M=4π2R3GT2        (I)

Here, M is the mass of the body around which another body revolves, R is the distance between the two bodies, G is the gravitational constant and T is the time period of body.

Conclusion:

Substitute 1350km for R, 6.67×1011Nm2/kg2 for G and 3.92days for T in equation (I).

  M=4π2(1350km(1000m1km))3(6.67×1011)(3.92days(24hr1day)(60min1hr)(60sec1min))2=9.713×10197.65kg=1.269×1019kg

Thus, the mass of the asteroid Electra is 1.269×1019kg .

(b)

To determine

The density of the asteroid Electra.

(b)

Expert Solution
Check Mark

Answer to Problem 38QP

The density of the asteroid Electra is 4.021×103kg/m3 .

Explanation of Solution

Write the expression for the density of the asteroid Electra.

  ρ=MV        (II)

Here, ρ is the density of the asteroid and V is the volume of the asteroid.

Write the expression for the volume of the asteroid.

  V=43πr3        (III)

Here, r is the radius of the asteroid.

The diameter of the asteroid is 182km. Therefore, the radius of the asteroid is 91km.

Conclusion:

Substitute 91km for r in equation (III).

  V=43π(91km(1000m1km))3=4.188(7.53×1014)m3=3.156×1015m3

Substitute 1.269×1019kg for M and 3.156×1015m3 for V in equation (II).

  ρ=1.269×1019kg3.156×1015m3=4.021×103kg/m3

Thus, the density of the asteroid Electra is 4.021×103kg/m3 .

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