Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 11, Problem 57P

(a)

To determine

The expression for the speed of each planet and for their relative speed.

(a)

Expert Solution
Check Mark

Answer to Problem 57P

The expression for the speed of planet 1 is m22Gd(m1+m2)_, planet 2 is m12Gd(m1+m2)_ and for their relative speed is 2G(m1+m2)d_.

Explanation of Solution

The system of two planets are isolated, so the energy as well as the momentum is conserved.

Write the expression for the conservation of energy.

    initial energy=final energyEi=Ef        (I)

Here, Ei is the initial energy, Ef is the final energy.

Write the expression for Ei.

    Ei=K1i+K2i+Ui        (II)

Here, K1i is the initial kinetic energy of planet 1, K2i is the initial kinetic energy of planet 2, Ui is the initial gravitational potential energy between the two planets

Write the expression for the Ef.

    Ef=K1f+K2f+Uf        (III)

Here, K1f is the final kinetic energy of planet 1, K2f is the kinetic energy of planet 2, , Uf is the final gravitational potential energy between the planets.

Use equation (II) and (III) in (I) to solve for the conservation of energy.

    K1i+K2i+Ui=K1f+K2f+Uf        (IV)

In the initial condition, the two planets are nearly at rest and they are infinite distance apart.

    K1i=0K2i=0Ui=0        (V)

Use equation (V) in (IV) to solve for the conservation of energy.

    0=K1f+K2f+Uf        (VI)

Write the expression for the conservation of momentum.

    (P1+P2)i=(P1+P2)f        (VII)

Here, P1i is the initial momentum of planet 1, P2i is the initial momentum of planet 2, P1f is the final momentum of planet 1, P2f is the final momentum of planet 2.

Use equation (V) in (VII) to solve for the conservation momentum.

    0=(P1+P2)f        (VIII)

Write the expression for the kinetic energy.

    K=12mv2        (IX)

Here, m is the mass, v is the velocity.

Write the expression for the momentum.

    P=mv        (X)

Write the expression for the Uf.

    Uf=Gm1m2d        (XI)

Here, G is the gravitational constant, d is the separation between the planets, m1 is the mass of planet 1, m2 is the mass of planet 2.

Use equation (IX) and (XI) in (VI) and it becomes,

    12m1v12+12m2v22Gm1m2d=012m1v12+12m2v22=Gm1m2dm1v12+m2v22=2Gm1m2d        (XII)

Use equation (X) in (VIII) to solve for v1.

    0=m1v1m2v2m1v1=m2v2v1=(m2m1)v2        (XIII)

Use equation (XIII) in (XII) to solve for v2.

    m2(m1m2v12+v22)=2Gm1m2dm1m2v12+v22=2Gm1m2d        (XIV)

Substitute the value of v1 with equation (XIII),

    m1m2(m2m1(v2))2+v22=2Gm1dm2m1v22+v22=2Gm1dv22(m2+m1m1)=2Gm1d        (XV)

Use equation (XV) to solve for v2.

    v2=m12Gd(m1+m2)        (XVI)

Use equation (XVI) in (XIII) to solve for v1.

    v1=m2m1(m12Gd(m1+m2))=m22Gd(m1+m2)        (XVII)

Write the expression for the relative velocity.

    vr=v1(v2)        (XVIII)

Here, vr is the relative velocity of the two planets.

Use equation (XVI) and (XVII) in (XVIII) to solve for vr.

    vr=(m22Gd(m1+m2))(m12Gd(m1+m2))=2G(m1+m2)d        (XIX)

Conclusion:

Therefore, the expression for the speed of planet 1 is m22Gd(m1+m2)_, planet 2 is m12Gd(m1+m2)_ and for their relative speed is 2G(m1+m2)d_.

(b)

To determine

The kinetic energy of the each planet just before the collision.

(b)

Expert Solution
Check Mark

Answer to Problem 57P

The kinetic energy of the planet 1 is 1.07×1032J_ and the planet 2 is 2.67×1031J_ just before the collision.

Explanation of Solution

Write the expression for the d.

    d=r1+r2        (XX)

Here, r1 is the radius of planet 1, r2 is the radius of planet 2.

Use equation (XVI) in (IX) to solve for K1.

    K1=12m1m22Gd(m1+m2)        (XXI)

Here, K1 is the kinetic energy of planet 1 just before the collision.

Use equation (XVII) in (IX) to solve for K2.

    K2=12m2m12Gd(m1+m2)        (XXII)

Here, K2 is the kinetic energy of planet 2 just before the collision.

Conclusion:

Substitute 3.00×106m for r1, 5.00×106m for r2 in equation (XX) to find d.

    d=(3.00×106m+5.00×106m)=8.00×106m

Substitute 6.67×1011Nm2/kg2 for G, 2.00×1024kg for m1, 8.00×1024kg for m2, 8.00×106m for d in equation (XXI) to find K1.

    K1=12(2.00×1024kg)(8.00×1024kg)2(6.67×1011Nm2/kg2)8.00×106m(2.00×1024kg+8.00×1024kg)=12(2.00×1024kg)(1.03×104m/s)=1.07×1032J

Substitute 6.67×1011Nm2/kg2 for G, 2.00×1024kg for m1, 8.00×1024kg for m2, 8.00×106m for d in equation (XXII) to find K2.

    K2=12(8.00×1024kg)(2.00×1024kg)2(6.67×1011Nm2/kg2)8.00×106m(2.00×1024kg+8.00×1024kg)=12(2.00×1024kg)(2.58×103m/s)=2.67×1031J

Therefore, the kinetic energy of the planet 1 is 1.07×1032J_ and the planet 2 is 2.67×1031J_ just before the collision.

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Principles of Physics: A Calculus-Based Text

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