Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305714892
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 13, Problem 13.58AP

(a)

To determine

The maximum height gain by the space vehicle.

(a)

Expert Solution
Check Mark

Answer to Problem 13.58AP

The maximum height gain by the space vehicle is vi2R2(gR12vi2).

Explanation of Solution

The initial speed of the vehicle is vi, the escape speed of space vehicle is vesc and the height of meteorite from the Earth’s surface is h.

Formula to calculate the maximum height gain by the space vehicle by the conservation of energy is,

    KE1+PE1=KE2+PE2        (1)

Here, KE1 is the kinetic energy of the space vehicle at the Earth’s surface, KE2 is the final kinetic energy of the space vehicle, PE1 is the potential energy of the space vehicle at the Earth’s surface and PE2 is final potential energy of the space vehicle.

 Formula to calculate the kinetic energy of the space vehicle at the Earth’s surface is,

    KE1=12mvi2

Here, m is the mass of the space vehicle and vi is the velocity of the space vehicle.

Formula to calculate the potential energy of the space vehicle at the Earth’s surface is,

    PE1=GMmR

Here, M is the mass of Earth, m is the mass of the space vehicle, R is the radius of the Earth and G is the universal gravitational constant.

Formula to calculate the potential energy space vehicle at the altitude is,

  PE2=GMmR+hmax

Here, hmax is the maximum height gain by the space vehicle.

The final kinetic energy of the space vehicle is zero because the space vehicle is rest at that point.

Substitute GMmR for PE1, GMmR+hmax for PE2, 12mvi2 for KE1 and 0 for KE2 in equation (1) to find hmax.

    12mvi2+(GMmR)=0+(GMmR+hmax)12mvi2=GMmRGMmR+hmax=GMm(1R1R+hmax)=GMm(hmaxR(R+hmax))        (2)

Further solve the above expression.

    12mvi2R(R+hmax)=GMmhmax12mvi2R2+12mvi2Rhmax=GMmhmaxhmaxm(GM12vi2R)=12mvi2R2hmax=vi2R22(GM12vi2R)

Write the expression for the acceleration due to gravity.

    g=GMR2GM=gR2

Here, g is the acceleration due to gravity.

Substitute gR2 for GM in the above equation to find hmax.

    hmax=vi2R22(gR212vi2R)=vi2R22R(gR12vi2)=vi2R2(gR12vi2)

Conclusion:

Therefore, the maximum height gain by the space vehicle is vi2R2(gR12vi2).

(b)

To determine

The speed of the meteorite to strike the Earth.

(b)

Expert Solution
Check Mark

Answer to Problem 13.58AP

The speed of the meteorite to strike the Earth is 2gh(1(1+hR)).

Explanation of Solution

The initial speed of the vehicle is vi, the escape speed of space vehicle is vesc and the height of meteorite from the Earth’s surface is h.

 From equation (2), the expression for the speed is given as,

    12mv2=GMm(hR(R+h))v2=2GMmm(hR(R+h))

Here, v is the speed of the meteorite to strike the Earth and h is the height of meteorite from the Earth’s surface.

Further solve the above expression.

    v2=2GMmm(hR(R+h))v=2GM(hR(R+h))

Substitute gR2 for GM in the above equation to find v.

    v=2gR2(hR(R+h))=2gR(h(R+h))=2gh(1(1+hR))

Conclusion:

Therefore, the speed of the meteorite to strike the Earth is 2gh(1(1+hR)).

(c)

To determine

The result from part (a) is consistent with h=ui2sin2θi2g

(c)

Expert Solution
Check Mark

Answer to Problem 13.58AP

The result from part (a) is consistent with h=ui2sin2θi2g.

Explanation of Solution

Consider a baseball is tossed up with an initial speed that is very small as compared to the escape speed.

    vi<vesc

Here, vesc is the escape speed of space vehicle.

As the initial speed that is very small. So the initial speed of the vehicle tends to be zero.

    vi0

From part (a), the maximum height gain by the space vehicle is,

    hmax=vi2R2(gR12vi2)

Substitute 0 for vi in the above equation to find hmax.

    hmax=vi2R2(gR0)=vi22g        (3)

Write the expression for the maximum height of the projectile motion of the baseball.

    h=vi2sin2θi2g

Here, h is the maximum height of the projectile motion of the baseball and θi is the angle of the projectile motion of the baseball.

From maximum height of the projectile motion of the baseball, the value of angle of the projectile motion of the baseball should be 90°.

Substitute 90° for θi in the above equation to find h.

    h=vi2sin290°2g=vi2(1)22g=vi22g        (4)

From equations (3) and (4).

    hmax=h

So, the maximum height gain by the space vehicle is consistent with h=ui2sin2θi2g.

Conclusion:

Therefore, the result from part (a) is consistent with h=ui2sin2θi2g.

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Chapter 13 Solutions

Bundle: Physics for Scientists and Engineers, Technology Update, 9th Loose-leaf Version + WebAssign Printed Access Card, Multi-Term

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