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A uniform, spherical, 1000.0-kg shell has a radius of 5.00 m. (a) Find the gravitational force this shell exerts on a 2.00-kg point mass placed at the following distances from the center of the shell: (i) 5.01 m, (ii) 4.99 m, (iii) 2.72 m. (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r = 0\) to \(r\) \(\rightarrow\) \(\infty\).

Short Answer

Expert verified
(i) \(5.30 \times 10^{-9}\) N, (ii) 0 N, (iii) 0 N. The force is zero inside the shell and decreases as 1/\(r^2\) outside.

Step by step solution

01

Understanding the Problem

We need to calculate the gravitational force between a uniform spherical shell and a point mass at varying distances. The shell's mass \(M\) is 1000.0 kg, and the point mass \(m\) is 2.00 kg. The shell's radius is 5.00 m. According to the shell theorem, if a point mass is outside a spherical shell, the shell can be treated as a point mass at its center. If it is inside, the gravitational force is zero.
02

Applying Newton's Gravity Law for Case (i)

For the point mass at a distance \(r = 5.01\) m, which is outside the shell, use the formula:\[F = \frac{G M m}{r^2}\]where \(G = 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2\). Substitute \(M = 1000.0\) kg, \(m = 2.00\) kg, and \(r = 5.01\) m.
03

Calculation for 5.01 m

Substitute the values into the gravitational force formula:\[F = \frac{(6.674 \times 10^{-11})(1000)(2.00)}{(5.01)^2}\]Calculate to find \(F \approx 5.30 \times 10^{-9}\) N.
04

Applying Shell Theorem for Case (ii)

For the point mass at 4.99 m, which is inside the shell, according to the shell theorem, the gravitational force is zero because the mass of the shell does not exert a net gravitational force within itself.
05

Applying Shell Theorem for Case (iii)

For the point mass at 2.72 m, which is also inside the shell, the gravitational force is again zero, per the shell theorem, as there is no net gravitational force exerted within the shell.
06

Sketching the Graph

Plot a graph where the x-axis represents the distance \(r\) from the center of the sphere, and the y-axis represents gravitational force \(F\). From \(r = 0\) to \(r = 5.00\) m, \(F = 0\). From \(r > 5.00\) m, \(F\) follows the curve \(F = \frac{G M m}{r^2}\), decreasing as \(r\) increases.

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental force that acts between two masses, attracting them towards each other. This force is described by Newton's law of universal gravitation, which we will discuss in detail later on. The key idea is that the gravitational force is always attractive and depends on both the masses involved and the distance between them.

In our original exercise, the gravitational force is calculated between a spherical shell and a point mass. The formula used to find this force is given by:
  • Force (\( F \)) is proportional to the product of the two masses, the shell (\( M \)) and the point mass (\( m \)).
  • This force is inversely proportional to the square of the distance (\( r \)) between the center of the spherical shell and the point mass.
It's important to note that gravitational force diminishes rapidly as the distance between the two masses increases, illustrating the inverse square law nature of this force.
Spherical Shell
A spherical shell is a three-dimensional shape that is hollow and defined by two radii, one outer and one possibly inner if it's a thick shell. In physics, particularly in relation to the gravitational force, spherical shells hold special significance due to the shell theorem derived from Newtonian gravity.

The shell theorem states:
  • Outside the shell: When a point mass is located outside a spherical shell, the shell’s gravitational force acts as though all its mass were concentrated at its center.
  • Inside the shell: If the point mass is inside the spherical shell, there is no net gravitational force exerted by the shell, regardless of the internal structure of the shell.
This property greatly simplifies calculations as it allows us to treat the shell as a point mass for external forces, providing that neat simplification evident in the given exercise solutions.
Newton's Law of Universal Gravitation
Newton's law of universal gravitation is a pivotal theory in physics that describes the attractive force between two bodies. It states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The magnitude of this force is given by the formula:\[F = \frac{G M m}{r^2}\]Here:
  • \( F \) is the gravitational force between the two masses.
  • \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \)), a constant of nature that quantifies the strength of gravity.
  • \( M \) and \( m \) are the masses of the two objects.
  • \( r \) is the distance between the center of the two masses.
This groundbreaking formula shows how gravitational force is universal across the cosmos, applying not just to celestial bodies like the moon and Earth but also to the interactions between everyday objects. It illustrates the direct proportionality to the product of the masses and the inverse square relationship with distance, a feature fundamental to understanding gravitational interactions.

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Most popular questions from this chapter

An experiment is performed in deep space with two uniform spheres, one with mass 50.0 kg and the other with mass 100.0 kg. They have equal radii, \(r =\) 0.20 m. The spheres are released from rest with their centers 40.0 m apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are 20.0 m apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the 50.0-kg sphere do the surfaces of the two spheres collide?

The star Rho\(^1\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho\(^1\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho\(^1\) Cancri?

Deimos, a moon of Mars, is about 12 km in diameter with mass 1.5 \(\times\) 10\(^{15}\) kg. Suppose you are stranded alone on Deimos and want to play a one- person game of baseball. You would be the pitcher, and you would be the batter! (a) With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it? Do you think you could actually throw it at this speed? (b) How long (in hours) after throwing the ball should you be ready to hit it? Would this be an action-packed baseball game?

Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about 200 km/s. (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzschild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?

Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 943.0 N on the earth weighs 915.0 N at the north pole of Planet X and only 850.0 N at its equator. The distance from the north pole to the equator is 18,850 km, measured along the surface of Planet X. (a) How long is the day on Planet X? (b) If a 45,000-kg satellite is placed in a circular orbit 2000 km above the surface of Planet X, what will be its orbital period?

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