Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 16, Problem 86P

(a)

To determine

The possible values of charges.

(a)

Expert Solution
Check Mark

Answer to Problem 86P

The charge of Sun can be 1.712×1020C and the charge of Earth can be 5.148×1014C.

Explanation of Solution

Write an expression for equilibrium condition for gravitational and electrical force.

  k|qS||qE|r2=GmSmEr2                                                                                                  (I)

Here, k is the Coulomb’s constant, qS is the charge of Sun, qE is the charge of Earth, r is the distance between Earth and Sun, G is the gravitational constant, mS  is the mass of Sun and mE is the mass of Earth.

Rewrite equation (I).

  |qS||qE|=GkmSmE                                                                                                   (II)

Consider,

  |qS|mS=|qE|mE                                                                                                               (III)

Rewrite equation (III) to find |qE|.

  |qE|=mEmS|qS|                                                                                                         (IV)

Rewrite equation (II) to find |qS|.

  |qS|=GkmSmE(1|qE|)                                                                                              (V)

Substitute equation (IV) in equation (V).

  |qS|=GkmSmE(1mEmS|qS|)=GkmSmE(mSmE|qS|)                                                                           (VI)

Rewrite equation (VI) to find |qS|.

  |qS|=GkmS                                                                                                         (VII)

Rewrite equation (III) to find |qS|.

  |qS|=mSmE|qE|                                                                                                       (VIII)

Rewrite equation (II) to find |qE|.

  |qE|=GkmSmE(1|qS|)                                                                                              (IX)

Substitute equation (VIII) in equation (IX).

  |qE|=GkmSmE(1mSmE|qE|)=GkmSmE(mEmS|qE|)                                                                          (X)

Rewrite equation (X) to find |qE|.

  qE=GkmE                                                                                                          (XI)

Conclusion:

Substitute 1.987×1030kg for mS, 6.674×1011Nm2/kg2 for G and 8.988×109Nm2/C2 for k in equation (VII) to find |qS|.

    |qS|=(6.674×1011Nm2/kg28.988×109Nm2/C2)(1.987×1030kg)=(0.74×1020C2/kg2)(1.987×1030kg)=1.712×1020C

Substitute 5.974×1024kg for mE, 6.674×1011Nm2/kg2 for G and 8.988×109Nm2/C2 for k in equation (XI) to find |qE|.

    |qE|=(6.674×1011Nm2/kg28.988×109Nm2/C2)(5.974×1024kg)=(0.74×1020C2/kg2)(5.974×1024kg)=5.148×1014C

Thus, the charge of Sun can be 1.712×1020C and the charge of Earth can be 5.148×1014C. Let the charge of Sun is positive and charge of Earth is negative.

(b)

To determine

If charge imbalance explain Earth’s orbit.

(b)

Expert Solution
Check Mark

Answer to Problem 86P

No, the charge imbalance explain Earth’s orbit.

Explanation of Solution

Consider the charge of electron and proton is different. The net charges on astronomical bodies are the sum of charge of the particles. If the magnitude of the charges of the proton and electron were not exactly equal, astronomical bodies will have a total charge with the same sign.

Like charges will repel each other. Since the charges of the astronomical bodies are similar, the force between them would be repulsive. The force responsible for the Earth’s orbit is attractive. Thus, the charge imbalance explain Earth’s orbit.

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Chapter 16 Solutions

Physics

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