COLLEGE PHYSICS-CONNECT ACCESS
COLLEGE PHYSICS-CONNECT ACCESS
5th Edition
ISBN: 9781260486834
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 6, Problem 57P
To determine

The speed of the rock.

Expert Solution & Answer
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Answer to Problem 57P

The speed of the rock is 3.3×104m/s_.

Explanation of Solution

The total energy of the system is conserved. That is the initial and final energies are equal.

    12mrockvi2GMEarthmrockdEarth(GMmoonmrockdmoon)=12mrockvf2GMEarthmrockh(GMmoonmrockRh)        (I)

Here, mrock is the mass of the rock, G is the gravitational constant, vi is the initial speed of the rock. vf is the final speed of the rock, MEarth is the mass of Earth, Mmoon is the mass of the moon, R is the distance between Earth and moon, h is the height of the rock above the Earth, dEarth is the distance to the Earth, and dmoon is the distance to the moon.

Add the term GMEarthmrockh in both sides of the equation (I), substrate GMmoonmrockRh from the both sides, and substitute, 0 for vi.

    12mrockvf2=12mrock×0GMEarthmrockdEarth+GMmoonmrockdmoonGMmoonmrockRh+GMEarthmrockhvf2=2G(MmoondmoonMEarthdEarthMmoonRh+MEarthh)        (II)

Given that, the gravitational forces on the rock from Earth and Moon are equal in magnitude and opposite direction.

    GMEarthmrockdEarth2=GMMoonmrockdMoon2dEarth2MEarth=dMoon2MMoon        (III)

Multiply both sides if equation (III) by MMoondMoon2.

    dEarth2MEarthMMoondMoon2=dMoon2MMoonMMoondMoon2dEarth=dMoonMEarthMMoon        (IV)

The distance between the Earth and moon is equal to the distance from the rock to the Earth plus the distance between the rock and the Moon.

    R=dEarth+dMoon        (V)

Use equation (IV) in (V).

    R=dMoonMEarthMMoon+dMoon=dMoon(1+MEarthMMoon)        (VI)

Similarly, R can be written as,

    R=dEarthMEarthMMoon+dEarth=dEarth(1+MEarthMMoon)        (VII)

Solve the equation (VII).

    dEarth=R1+MEarthMMoon        (VIII)

Similarly dmoon can be written as,

    dMoon=R1+MEarthMMoon        (IX)

Conclusion:

Substitute, 5.974×1024kg for MEarth, 7.349×1022kg for Mmoon, and 3.845×108m for R in the equation (VIII).

    dMoon=3.845×108m1+5.974×1024kg7.349×1022kg=0.3845×108m

Substitute, 5.974×1024kg for MEarth, 7.349×1022kg for Mmoon, 6.674×1011Nm2/kg2 for G, 0.3845×108m for dMoon,dEarth, 7.2×105m for h, and 3.845×108m for R in the equation (II).

    vf2=2(6.674×1011Nm2/kg2)×(7.349×1022kg0.3845×108m5.974×1024kg0.3845×108m7.349×1022kg3.845×108m7.2×105m+5.974×1024kg7.2×105m)=3.3×104m/s

Therefore, the speed of the rock is 3.3×104m/s_.

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Chapter 6 Solutions

COLLEGE PHYSICS-CONNECT ACCESS

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