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(a) If the legendary apple of Newton could be released from rest at a height of \(2 \mathrm{~m}\) from the surface of a neutron star with a mass 1.5 times that of our Sun and a radius of \(20 \mathrm{~km},\) what would be the apple's speed when it reached the surface of the star? (b) If the apple could rest on the surface of the star, what would be the approximate difference between the gravitational acceleration at the top and at the bottom of the apple? (Choose a reasonable size for an apple; the answer indicates that an apple would never survive near a neutron star.)

Short Answer

Expert verified
(a) The speed would be extremely high. (b) The gravitational difference would be enormous, destroying the apple.

Step by step solution

01

Understand the Problem

We are given the height from which an apple is released and the characteristics of a neutron star. We need to find the apple's speed when it reaches the surface and the difference in gravitational acceleration at the apple's top and bottom.
02

Calculate the Gravitational Force (a)

The gravitational constant is given by Newton's law of gravitation: \[ F = \frac{G M m}{r^2} \] where \( M \) and \( m \) are the masses of the neutron star and the apple, respectively; \( r \) is the distance from the center of the star to the apple, and \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2 \).
03

Apply Conservation of Energy for Part (a)

The apple initially has gravitational potential energy, which is converted to kinetic energy as it falls: \[ mgh = \frac{1}{2} mv^2 \] where \( h = 2 \) m is the height and \( g \) is the gravitational acceleration just above the neutron star surface, calculated from \( g = \frac{G M}{R^2} \). Since the apple is released from rest, its initial kinetic energy is zero.
04

Calculate Apple's Speed Using Energy Conservation (a)

First, find the gravitational acceleration \( g \) exactly at the surface using:\[ g = \frac{G M}{R^2} = \frac{(6.674 \times 10^{-11}) (1.5 \times 2 \times 10^{30})}{(20,000)^2} \].Calculate \( v \) using:\[ v = \sqrt{2gh} \].
05

Determine Difference in Gravitational Acceleration for Part (b)

Assume the apple's height is about \( 8 \) cm or \( 0.08 \) m. Calculate the gravitational acceleration at the top and bottom of the apple using:\[ g_{top} = \frac{G M}{(R + 0.08)^2} \] and\[ g_{bottom} = \frac{G M}{R^2} \].The difference in acceleration is \( \Delta g = g_{top} - g_{bottom} \).
06

Evaluate the Safety of the Apple (b)

With the calculated \( \Delta g \), we can see that the stresses from the gravitational difference would be immense, causing the apple to disintegrate, confirming the apple couldn't survive near the neutron star.

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Key Concepts

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

Neutron Star
Neutron stars are fascinating cosmic objects. They are the remnants of massive stars that have exploded as supernovae. Post-explosion, what's left is incredibly dense—so dense that a sugar-cubed size of neutron star material would weigh as much as a mountain on Earth.
A neutron star typically has a mass greater than that of our Sun, yet it is compressed into a radius of around 10 to 20 kilometers. Imagine squeezing something several times the mass of the Sun into an area just a few kilometers across!
This density results in an extreme gravitational pull, making neutron stars one of the most compelling subjects in astrophysics. They provide insights into not only stellar evolution but also extreme states of matter that cannot be replicated in any laboratory on Earth.
Gravitational Acceleration
Gravitational acceleration is the rate at which gravity causes an object to speed up as it falls towards a massive body, like a planet or star. It is a ubiquitous force and is calculated using the formula: \[ g = \frac{G M}{R^2} \]where:
  • \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
  • \( M \) is the mass of the astronomical object
  • \( R \) is the distance from this body's center to the object's location
On Earth, this value is about \(9.81 \, \text{m/s}^2\).
However, near a neutron star, this acceleration is overwhelmingly higher due to its immense mass packed into a small volume. Even at a mere two meters above a neutron star’s surface, an apple would experience extreme acceleration. Hence, when talking about neutron stars, gravitational acceleration plays a pivotal role in understanding their powerful effects.
Conservation of Energy
The principle of conservation of energy is fundamental in physics. It states that energy in a closed system remains constant; it can transform from one form to another but cannot be created or destroyed.
In the context of the apple falling towards a neutron star, initially, it holds gravitational potential energy due to its height. As it falls, this energy converts to kinetic energy, increasing its speed. This transformation is described by the equation:\[ mgh = \frac{1}{2} mv^2 \]where:
  • \( m \) is the mass of the apple
  • \( g \) is the gravitational acceleration
  • \( h \) is the height from which it falls
  • \( v \) is the speed of the apple just before it impacts
Gravitational potential energy at height converts seamlessly into kinetic energy just above the surface, showcasing nature's efficiency. In scenarios involving immense gravitational fields, such as those around neutron stars, conservation of energy aids in calculating potentially significant speeds, as observed in the exercise.
Newton's Law of Gravitation
Newton's Law of Gravitation provides a foundational understanding of how gravity acts between two masses. Expressed as: \[ F = \frac{G M m}{r^2} \]it explains how the force (\( F \)) of gravity between two objects is directly proportional to the product of their masses (\( M \) and \( m \)) and inversely proportional to the square of the distance \( r \) between their centers.
This law is crucial for calculating the gravitational forces encountered by objects in space, from apples to astronauts, offering critical insights into how objects interact with each other gravitationally.
In the case of neutron stars, their massive gravitational pull can be understood using this law, clarifying how they can exert such powerful forces even on objects as small as an apple. Newton's gravitation law remains as relevant today as during Newton's time, underpinning countless discoveries in astrophysics.

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Most popular questions from this chapter

In a shuttle craft of mass \(m=3000 \mathrm{~kg}\), Captain Janeway orbits a planet of mass \(M=9.50 \times 10^{25} \mathrm{~kg},\) in a circular orbit of radius \(r=4.20 \times 10^{7} \mathrm{~m} .\) What are (a) the period of the orbit and (b) the speed of the shuttle craft? Janeway briefly fires a forwardpointing thruster, reducing her speed by \(2.00 \%\). Just then, what are (c) the speed, (d) the kinetic energy, (e) the gravitational potential energy, and (f) the mechanical energy of the shuttle craft? (g) What is the semimajor axis of the elliptical orbit now taken by the craft? (h) What is the difference between the period of the original circular orbit and that of the new elliptical orbit? (i) Which orbit has the smaller period?

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