EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 11, Problem 106P

(a)

To determine

Toshow: The ratio of the force exerted on a point particle on the surface of Earth by the Sun to that exerted by the moon is, Msrm2/Mmrs2 .

(a)

Expert Solution
Check Mark

Explanation of Solution

Given information :

Mass of the Sun =Ms

Mass of the Moon =Mm

Distance of the particle from Earth to Sun =rs

Distance of the particle from Earth to Moon =rm

Mass of the Sun =1.99×1030kg

Mass of the Moon =7.36×1022kg

Average orbital distance between earth and sun =1.50×1011m

Average orbital distance between earth and moon =3.84×108m

Formula used :

Gravitational force between two masses ( M and m ) separated by distance r is:

  F=GMmr2

G is the gravitational constant.

Proof:

Express the force exerted by the sun on a body of water of mass, m ,

  Fs=GMsmrs2...(1)

Express the force exerted by the moon on a body of water of mass m ,

  Fm=GMmmrm2...(2)

Divide equation (1) by (2)

  FsFm=Msrm2Mmrs2

Substitute the values and solve:

  FsFm=(1.99×1030kg)(3.84×108m)2(7.36×1022kg)(1.50×1011m)2FsFm=177

(b)

To determine

ToShow: dF/F=(2dr)/r

(b)

Expert Solution
Check Mark

Explanation of Solution

Given information :

Mass of the Sun =Ms

Mass of the Moon =Mm

Distance of the particle from Earth to Sun =rs

Distance of the particle from Earth to Moon =rm

Formula used :

Gravitational force between two masses ( M and m ) separated by distance r is:

  F=GMmr2

G is the gravitational constant.

  

Calculation:

  F=Gm1m2r2

Differentiate the expression with respect to r :

  dFdr=2Gm1m2r3dFdr=2Fr

Re-arrange the expression:

  dFF=2drr

(c)

To determine

To Show:For a small difference in distance compared to the average distance, the ratio of the differential gravitational force exerted by the Sun to the differential gravitational force exerted by the moon on Earth’s oceans is given by

  ΔFsΔFm(Msrm3)(Mmrs3) and to calculate the ratio.

(c)

Expert Solution
Check Mark

Explanation of Solution

Given information :

Mass of the Sun =Ms

Mass of the Moon =Mm

Distance of the particle from Earth to Sun =rs

Distance of the particle from Earth to Moon =rm

Formula used :

From part b, ΔF=2FrΔr

Calculations:

The change in force ΔF for a small change in distance Δr .

  ΔF=2FrΔr

  ΔFs=2GmMsrs2rsΔrsΔFs=2GmMsrs3Δrs

  ΔFm=2GmMmrm2Δrm

  ΔFsΔFm=Msrs3ΔrsMmrm3ΔrmΔFsΔFm=Msrm3ΔrsMmrs3Δrm

Because the difference in the distance is small, Δrs and Δrm can be neglected.

  ΔFsΔFmMsrm3Mmrs3

Substitute the numerical values

  ΔFsΔFm=(1.99×1030kg)(3.84×108m)3(7.36×1022kg)(1.50×1011m)3ΔFsΔFm=0.454

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Chapter 11 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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