College Physics, Volume 1
College Physics, Volume 1
2nd Edition
ISBN: 9781133710271
Author: Giordano
Publisher: Cengage
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Chapter 5, Problem 70P

(a)

To determine

The acceleration due to gravity at the surface of the Vesta.

(a)

Expert Solution
Check Mark

Answer to Problem 70P

The acceleration due to gravity at the surface of the Vesta is 0.293m/s2.

Explanation of Solution

Write the expression from the kinematics equation of motion.

  Δy=v0t+12at2        (I)

Here, Δy is the height of the rock drops, v0 is the initial speed of the rock, t is the time taken by the rocks strikes on the ground, and a is the acceleration.

Substitute 0 for v0 in above relation and rewrite for acceleration.

  Δy=(0)t+12at2Δy=12at2a=2Δyt2

Conclusion:

Substitute 1.5m for Δy and 3.2s for t in above relation to find acceleration..

  a=2(1.5m)(3.2s)2=0.293m/s2

Therefore, the acceleration due to gravity at the surface of the Vesta is 0.293m/s2.

(b)

To determine

The mass of the Vesta.

(b)

Expert Solution
Check Mark

Answer to Problem 70P

The mass of the Vesta is 3.2×1014kg.

Explanation of Solution

Write the expression for acceleration due to gravity.

  g=GMr2        (II)

Here, M is the mass of the Vesta, G is the gravitational constant, r is the radius of the surface of Vesta, and g is the acceleration due to gravity.

Rewrite the above expression for mass.

  M=gr2G        (III)

Conclusion:

Substitute 0.293m/s2 for g, 6.67×1011Nm2/kg2 for G, and 270m for r in equation (III).

  M=(0.293m/s2)(270m)2(6.67×1011Nm2/kg2)=3.2×1014kg

Therefore, the mass of the Vesta is 3.2×1014kg.

(c)

To determine

The height of the astronaut could jump on Vesta planet.

(c)

Expert Solution
Check Mark

Answer to Problem 70P

The height of the astronaut could jump on Vesta planet is 27.4m.

Explanation of Solution

Write the expression from the equation of motion.

  v2=v022gΔy        (IV)

Here, v is the final speed, v0 is the initial speed, Δy is the height, and g is the acceleration due to gravity.

Rewrite the above expression for initial speed by substituting 0 for v.

  v0Earth2=2gEarthΔyEarth        (V)

Here, v0Earth is the initial speed of the object on Earth, ΔyEarth is the height of the rock drops on Earth surface, and gEarth is the acceleration due to gravity on Earth surface.

Similarly the initial speed of the object on Vesta surface,

  v0Vesta2=2gVestaΔyVesta        (VI)

Here, v0Vesta is the initial speed of the object on Vesta, ΔyVesta is the height of the rock drops on Vesta surface, and gVesta is the acceleration due to gravity on Vesta surface.

Conclusion:

Compare the equations (V) and (VI).

  v0Earth2=v0Vesta22gEarthΔyEarth=2gVestaΔyVestagEarthΔyEarth=gVestaΔyVestaΔyVesta=(gEarthgVesta)ΔyEarth

Substitute 0.293m/s2 for gVesta, 9.80m/s2 for gEarth, and 82cm for ΔyEarth in above relation to find ΔyVesta.

  ΔyVesta=(9.80m/s20.293m/s2)(82cm)(0.01m1cm)=27.4m

Therefore, the height of the astronaut could jump on Vesta planet is 27.4m.

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Chapter 5 Solutions

College Physics, Volume 1

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