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Chapter 13: Problem 35

The acceleration due to gravity at the north pole of Neptune is approximately 10.7 \(\mathrm{m} / \mathrm{s}^{2} .\) Neptune has mass \(1.0 \times 10^{26} \mathrm{kg}\) and radius \(2.5 \times 10^{4} \mathrm{km}\) and rotates once around its axis in about 16 \(\mathrm{h}\) . (a) What is the gravitational force on a 5.0 -kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

Short Answer

Expert verified
(a) 53.5 N. (b) 52.984 N.

Step by step solution

01

Determine Gravitational Force at North Pole

To find the gravitational force on a 5.0 kg object, use the formula for weight, which is the product of mass and gravitational acceleration: \( F = m \cdot g \). Here, \( m = 5.0 \, \text{kg} \) and \( g = 10.7 \, \text{m/s}^2 \). So, \( F = 5.0 \, \text{kg} \times 10.7 \, \text{m/s}^2 = 53.5 \, \text{N} \).
02

Determine Centripetal Acceleration at Equator

First, convert Neptune's rotational time into seconds: \( 16 \, \text{h} \times 3600 \, \text{s/h} = 57600 \, \text{s} \). Next, convert Neptune's radius to meters by multiplying by 1000: \( 2.5 \times 10^{4} \, \text{km} = 2.5 \times 10^{7} \, \text{m} \). Use the formula for centripetal acceleration \( a_c = \frac{4 \pi^2 R}{T^2} \), where \( R \) is the radius and \( T \) is the period in seconds.
03

Calculate Centripetal Acceleration

Substitute the values into the centripetal acceleration formula: \( a_c = \frac{4 \pi^2 \times 2.5 \times 10^{7} \, \text{m}}{(57600 \, \text{s})^2} \). Calculation gives \( a_c \approx 0.1032 \, \text{m/s}^2 \).
04

Determine Apparent Weight at Equator

The apparent weight of an object considers both gravitational force and the centripetal force. Calculate using \( F_{apparent} = F_{gravity} - F_{centripetal} \), where \( F_{centripetal} = m \cdot a_c \). \( F_{centripetal} = 5.0 \, \text{kg} \times 0.1032 \, \text{m/s}^2 \approx 0.516 \, \text{N} \).
05

Calculate Resultant Apparent Weight

Subtract the centripetal force from the gravitational force: \( F_{apparent} = 53.5 \, \text{N} - 0.516 \, \text{N} \approx 52.984 \, \text{N} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force Calculation
Gravitational force is the force with which a planet attracts an object towards its center. Calculating this force requires knowing both the mass of the object and the gravitational acceleration of the planet. Use the formula:
  • \( F = m \cdot g \)
where \( F \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity.
For Neptune, the north pole's gravitational acceleration is approximately \( 10.7 \mathrm{m/s}^2 \). Given a 5.0 kg object, insert these values into the formula to compute the force.
Thus, \( F = 5.0 \cdot 10.7 = 53.5 \) N. This is the force exerted by Neptune on the object. It's a straightforward calculation that combines understanding of gravity with simple multiplication.
Centripetal Force
Centripetal force is what keeps an object moving in a circular path. On a rotating planet like Neptune, objects at the equator experience this force due to the planet's rotation.
To determine the centripetal acceleration, use the formula:
  • \( a_c = \frac{4 \pi^2 R}{T^2} \)
where \( R \) is the planet's radius converted into meters and \( T \) is the rotation period in seconds.
For Neptune, convert its radius from \( 2.5 \times 10^{4} \) km to meters and rotation period from 16 hours to seconds. Plug these into the equation:
\( R = 2.5 \times 10^7 \) m and \( T = 57600 \) s.
This provides a centripetal acceleration \( a_c \approx 0.1032 \mathrm{m/s}^2 \).
The centripetal force \( F_c \) on a 5.0 kg object is then calculated by \( F_c = m \cdot a_c \approx 0.516 \) N, demonstrating how rotational motion influences apparent weight.
Apparent Weight
Apparent weight differs from actual weight due to external factors like rotation. For an object on Neptune's equator, its weight is reduced by the centripetal force caused by the planet's rotation.
Use the formula for apparent weight:
  • \( F_{apparent} = F_{gravity} - F_{centripetal} \)
where \( F_{gravity} \) is the gravitational force and \( F_{centripetal} \) is the force due to rotation.
For a 5.0 kg object, the gravitational force \( F_{gravity} = 53.5 \) N, and \( F_{centripetal} \approx 0.516 \) N.
Calculate the apparent weight as:
\( F_{apparent} = 53.5 - 0.516 = 52.984 \) N.
This value shows that the object feels slightly lighter due to Neptune's rapid rotation, which decreases the force felt by an object spinning along its equator.
Planetary Rotation
Planetary rotation influences various physical phenomena on the planet's surface. It affects the apparent weight of objects and contributes to the centripetal force they experience.
Neptune rotates once every 16 hours, translating to a period \( T = 57600 \) seconds. This rapid spin impacts all equatorial objects:
  • Reduces their apparent weight due to the centrifugal effect
  • Contributes to the centripetal acceleration
By affecting gravitational calculations, rotation is a crucial factor in understanding how objects behave on spinning bodies.
Comparing to Neptune's poles where there's minimal rotation influence, the impact is evident at the equator. Therefore, rotational dynamics are essential for comprehensive planetary analysis.

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Most popular questions from this chapter

Kirkwood Gaps. Hundreds of thousands of asteroids orbit the sun within the asteroid belt, which extends from about \(3 \times 10^{8} \mathrm{km}\) to about \(5 \times 10^{8} \mathrm{km}\) from the sun. (a) Find the orbital period (in years) of (i) an asteroid at the inside of the belt and (ii) an asteroid at the outside of the belt. Assume circular orbits. (b) In 1867 the American astronomer Daniel Kirkwood pointed out that several gaps exist in the asteroid belt where relatively few asteroids are found. It is now understood that these Kirkwood gaps are caused by the gravitational attraction of Jupiter, the largest planet, which orbits the sun once every 11.86 years. As an example, if an asteroid has an orbital period half that of Jupiter, or 5.93 years, on every other orbit this asteroid would be at its closest to Jupiter and feel a strong attraction toward the planet. This attraction, acting over and over on successive orbits, could sweep asteroids out of the Kirkwood gap. Use this hypothesis to determine the orbital radius for this Kirkwood gap. (c) One of several other Kirkwood gaps appears at a distance from the sun where the orbital period is 0.400 that of Jupiter. Explain why this happens, and find the orbital radius for this Kirkwood gap.

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