PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 11, Problem 80P

(a)

To determine

To Calculate:The value of constant

(a)

Expert Solution
Check Mark

Answer to Problem 80P

  6.4×108kg/m2

Explanation of Solution

Given data:

The density of the sphere, ρ(r)=Cr

Radius, =5.0m

Mass of the sphere, M=1.0×1011kg

Formula Used:

Mass = Density × Volume

Calculation:

The density of the sphere is

  ρ(r)=Cr...........(1)

Here C is the constant and r is the distance.

The density of the sphere is varied by a distance so the differential element of the sphere is ,

  dm=ρdV=ρ(4πr2dr)

Integrate within the limits 0 to R .

  M=(Cr)0R(4πr2dr)=4πC05.0mrdr=4πC[r22]05.0=πC(50m)2

Therefore the constant C is,

  C=Mπ(50m2)

Substitute 1.0×1011kg for M and solve for C .

  C=(1.0×1011kg)π(50m2)=6.4×108kg/m2

Conclusion:

The constant C is 6.4×108kg/m2 .

(b)

To determine

The acceleration due to gravity for a distance r>5.0m .

The gravitation field with in the region r>5.0m .

(b)

Expert Solution
Check Mark

Answer to Problem 80P

The acceleration due to gravity for a distance r>5.0m is 6.7N×m2/kg2r2 .

The gravitation field with in the region r<5.0m is 0.27 N/kg .

Explanation of Solution

Given data:

The density of the sphere, ρ(r)=Cr

Radius, =5.0m

Mass of the sphere, M=1.0×1011kg

The constant C is 6.4×108kg/m2 .

Formula used:

Gravitational field:

  g=GMr2r^

Here, G is the gravitational constant, M is the mass and r is the distance of the point from the center of the sphere.

Calculation:

The expression for the magnitude of gravitational field at a point outside (r>5.0m) the sphere is,

  g=GMr2

Substitute 6.673×1011Nm2/kg2forG,1.0×1011kg for M and solve for g .

  g=(6.673×1011Nm2/kg2)(1.0×1011kg)r2=6.7N×m2/kg2r2

Therefore, the acceleration due to gravity for a distance r>5.0m is 6.7N×m2/kg2r2

The expression for the gravitational field at a point inside (r<5.0m) the sphere is,

  g=GMr2r^

Since the density of the sphere is varying with the distance, so gravitational field is given by for

  r<5.0m

  g=G0rρ(4πr2)drr2=G0r(Cr)(4πr2)drr2=G4πC0rrdrr2=G4πC(r22)r2=2πGC

Substitute 6.673×1011Nm2/kg2 for G, 6.4×108kg/m2 for C and solve for g .

  g=2πGC=2π(6.673×1011N×m2/kg2)(6.4×108kg/m2)=0.27N/kg

Conclusion:

The acceleration due to gravity for a distance r>5.0m is 6.7N×m2/kg2r2 .

The gravitation field with in the region r<5.0m is 0.27 N/kg .

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Chapter 11 Solutions

PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS

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