COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 97P
To determine

The minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls.

Expert Solution & Answer
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Answer to Problem 97P

The minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls is 617.9 km/s .

Explanation of Solution

The total mechanical energy of the object will be constant. However, after the object escapes the both Earth’s and the Sun’s gravitational pulls, its total energy will be zero.

Write the equation for the conservation of energy of the object.

  Ki+Ui=0        (I)

Here, Ki is the initial kinetic energy of the object and Ui is the initial potential energy of the object.

Write the equation for Ki .

  Ki=12mvi2        (II)

Here, m is the mass of the object and vi is the minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls

Write the equation for Ui .

  Ui=UiS+UiE        (III)

Here, UiS is the gravitational pull of the Sun on the object and UiE is the gravitational pull of the Earth on the object.

Write the equation for UiS.

  UiS=GMSmRS        (IV)

Here, G is the universal gravitation constant, MS is the mass of the sun and RS is the radius of the Sun.

Write the equation for UiE.

  UiE=GMEmRE        (V)

Here, ME is the mass of the Earth and RE is the distance from the sun to the Earth.

Put equations (IV) and (V) in equation (III).

  Ui=GMSmRS+(GMEmRE)=GMSmRSGMEmRE=Gm(MSRS+MERE)        (VI)

Put equations (II) and (VI) in equation (I).

  12mvi2+[Gm(MSRS+MERE)]=012mvi2Gm(MSRS+MERE)=0vi2=2Gmm(MSRS+MERE)vi=2G(MSRS+MERE)        (VII)

Conclusion:

The value of G is 6.67×1011 m3/kgs2 , the mass of the sun is 1.989×1030 kg , the radius of the sun is 6.95×105 km , the mass of earth is 5.972×1024 kg and the distance from the sun to the earth is 149.6×106 km .

Substitute 6.67×1011 m3/kgs2 for G , 1.989×1030 kg for MS , 6.95×105 km for RS , 5.972×1024 kg for ME and 6.37×103 km for RE in equation (VII) to find vi.

  vi=2(6.67×1011 m3/kgs2)(1.989×1030 kg6.95×105 km1000 m1 km+5.972×1024 kg149.6×106 km1000 m1 km)=617.9×103 m/s=617.9×103 m/s1 km1000 m=617.9 km/s

Therefore, the minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls is 617.9 km/s.

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Chapter 6 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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