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You are sitting in the front row in your physics class. Estimate the gravitational force that the instructor exerts on you. Identify the assumptions and approximations you made to reach your answer. How does the magnitude that you estimate for this force compare with the gravity force exerted on you by the earth?

Short Answer

Expert verified
First, we get the gravitational force between student and the instructor to be approximately 2.52 x 10^-8 N. Second, the gravitational force between the student and the earth will be around 490 N. The gravitational force between the Earth and the student is significantly larger, approximately 1.94 x 10^10 times bigger than the force between the student and the instructor.

Step by step solution

01

Calculating the gravitational force between the student and the instructor

Using the formula for gravitational force, F = G*m1*m2/r^2, where G is the gravitational constant (6.674 x 10^-11), m1 and m2 are the masses of the student and the teacher and r is the distance between them. After substituting the values we will get the gravitational force between student and the instructor.
02

Calculating the gravitational force between the student and the earth

Using the same formula, but now m1 is the mass of the student and m2 is the mass of the Earth, and r is the distance from the student to the center of the Earth. Substituting the values we will get the gravitational force between student and the earth.
03

Comparing the gravitational forces

By comparing the two gravitational forces calculated in step 1 and step 2, we can then find out how the magnitudes differ.

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Key Concepts

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

Gravitational Constant
The gravitational constant, denoted by the symbol G, is a key figure in the calculation of gravitational force between two masses. Its value is approximately \( 6.674 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2} \). This might seem like an inconceivably small number, and indeed it is. This is why we can walk past buildings and other large objects without feeling any gravitational pull. Despite its small size, the gravitational constant is fundamental in understanding the strength of gravity between any two objects in the universe, no matter their distance.

When estimating the gravitational force between you and your physics instructor, this constant plays a crucial role. Without it, we couldn't calculate the invisible string that physically ties all matter together. The constant G ensures that we have a standardized number which doesn't change, whether we're calculating the pull between a student and a teacher or the Earth and the Moon.
Newton's Law of Universal Gravitation
Issac Newton's law of universal gravitation is a fundamental principle that tells us every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically speaking, this is represented as \( F = G\frac{m_1 \times m_2}{r^2} \), where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of mass of the two objects.

In the classroom scenario, applying Newton's law lets you estimate the force by substituting in the masses of yourself and your instructor and the distance between you two. It's important to note that the law applies to point masses, so the assumption here is that the mass of yourself and your instructor can each be concentrated at a point. Although this isn't strictly true, it is a reasonable approximation for such calculations and simplifies the problem considerably.
Force Comparison
When we compare gravitational forces, we're often struck by the vast differences in magnitude. In your physics class example, the gravitational force between you and your instructor is extraordinarily smaller compared to the gravitational force exerted on you by the Earth. This is primarily due to the huge disparity in masses and the distance involved.

Using Newton’s law of universal gravitation, the force between smaller objects, such as a student and a teacher, is minuscule because both the masses and the distance between them are comparatively small. However, the mass of the Earth is so significant, and despite the larger distance from the student to the center of the Earth, the resultant force is what we experience as our weight.
  • The mass of the Earth largely overshadows other influences,
  • Distance plays a critical role, but the squared relationship means that, even with further distances, massive objects like Earth can exert a considerable force.
In essence, the force comparison serves as a remarkable demonstration of the universality of gravity: it’s always there, but its effects range dramatically based on mass and distance.

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Most popular questions from this chapter

The mass of Venus is \(81.5 \%\) that of the earth, and its radius is \(94.9 \%\) that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs \(75.0 \mathrm{~N}\) on earth, what would it weigh at the surface of Venus?

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at \(30,000 \mathrm{~km} / \mathrm{s}\). (a) How far are these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass. (c) What is the radius of its event horizon?

Aura Mission. On July 15, 2004 NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit \(705 \mathrm{~km}\) above the earth's surface. Assume a circular orbit. (a) How many hours does it take this satellite to make one orbit? (b) How fast (in \(\mathrm{km} / \mathrm{s})\) is the Aura spacecraft moving?

A narrow uniform rod has length \(2 a\). The linear mass density of the rod is \(\rho,\) so the mass \(m\) of a length \(l\) of the rod is \(\rho l\). (a) A point mass is located a perpendicular distance \(r\) from the center of the rod. Calculate the magnitude and direction of the force that the rod exerts on the point mass. (Hint: Let the rod be along the \(y\) -axis with the center of the rod at the origin, and divide the rod into infinitesimal segments that have length \(d y\) and that are located at coordinate \(y\). The mass of the segment is \(d m=\rho d y\). Write expressions for the \(x\) - and \(y\) -components of the force on the point mass, and integrate from \(-a\) to \(+a\) to find the components of the total force. Use the integrals in Appendix B.) (b) What does your result become for \(a \gg r ?\) (Hint: Use the power series for \((1+x)^{n}\) given in Appendix B.) (c) For \(a \gg r,\) what is the gravitational field \(g=\boldsymbol{F}_{g} / m\) at a distance \(r\) from the rod? (d) Consider a cylinder of radius \(r\) and length \(L\) whose axis is along the rod. As in part (c), let the length of the rod be much greater than both the radius and length of the cylinder. Then the gravitational ficld is constant on the curved side of the cylinder and perpendicular to it, so the gravitational flux \(\Phi_{g}\) through this surface is cqual to \(g A\), where \(A=2 \pi r L\) is the area of the curved side of the cylinder (see Problem 13.59 ). Calculate this flux. Write your result in terms of the mass \(M\) of the portion of the rod that is inside the cylindrical surface. How does your result depend on the radius of the cylindrical surface?

The dwarf planet Pluto has an elliptical orbit with a semi-major axis of \(5.91 \times 10^{12} \mathrm{~m}\) and eccentricity \(0.249 .\) (a) Calculate Pluto's orbital period. Express your answer in seconds and in earth years. (b) During Pluto's orbit around the sun, what are its closest and farthest distances from the sun?

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