   # As thermonuclear fusion proceeds in its core, the Sun loses mass at a rate of 3.64 × 10 9 kg/s. During the 5 000-yr period of recorded history, by how much has the length of the year changed due to the loss of mass from the Sun? Suggestions: Assume the Earth’s orbit is circular. No external torque acts on the Earth–Sun system, so the angular momentum of the Earth is constant. ### Physics for Scientists and Enginee...

9th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305116399

#### Solutions

Chapter
Section ### Physics for Scientists and Enginee...

9th Edition
Raymond A. Serway + 1 other
Publisher: Cengage Learning
ISBN: 9781305116399
Chapter 13, Problem 13.77AP
Textbook Problem
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## As thermonuclear fusion proceeds in its core, the Sun loses mass at a rate of 3.64 × 109 kg/s. During the 5 000-yr period of recorded history, by how much has the length of the year changed due to the loss of mass from the Sun? Suggestions: Assume the Earth’s orbit is circular. No external torque acts on the Earth–Sun system, so the angular momentum of the Earth is constant.

To determine

The length of the year changed due to the loss of mass from the Sun.

### Explanation of Solution

The Sun loses mass at a rate of 3.64×1009kg/s in a 5000yr period of recorded history.

Formula to calculate the velocity of the Earth during circular motion is,

v=2πRT        (I)

Here, R is the radius of the orbit, v is the velocity of the Earth and T is the time taken by Earth to complete one revolution.

The Earth is revolving around the Sun, so the gravitational force is balanced by the centrifugal force.

GMMER2=MEv2R

Here, G is the universal gravitational constant and ME is the mass of the Earth.

Substitute 2πRT for v in above expression.

GMMER2=ME(2πRT)2RGMT2=4π2R3        (II)

Formula to calculate the linear momentum of the Earth is ,

L=MEvR

Here, L is the linear momentum.

Substitute 2πRT for v in above expression.

L=ME2πRTRR2=LT2πMER=(LT2πME)12

Substitute (LT2πME)12 for R in equation (II).

GMT2=4π2((LT2πME)12)3MT12=4π2G(L2πME)32

The mass of the Earth is constant; linear momentum is constant and G is also constant so the whole term 4π2G(L2πME)32 is constant

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