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Chapter 14: Problem 11

The masses and coordinates of three spheres are as follows: \(20 \mathrm{~kg}, x=0.50 \mathrm{~m}, y=1.0 \mathrm{~m} ; 40 \mathrm{~kg}\), \(x=-1.0 \mathrm{~m}, y=-1.0 \mathrm{~m} ; 60 \mathrm{~kg}, x=0 \mathrm{~m}, y=-0.50 \mathrm{~m} .\) What is the magnitude of the gravitational force on a \(20 \mathrm{~kg}\) sphere located at the origin due to the other spheres?

Short Answer

Expert verified
The magnitude of the gravitational force on the 20 kg sphere at the origin is approximately 2625 G.

Step by step solution

01

- Identify the Forces

Identify the gravitational forces between the 20 kg sphere at the origin and the other two spheres (40 kg and 60 kg). Use Newton's Law of Gravitation: Newton's Law of Gravitation: F = G frac {{m1 m2}}{{r^2}} Where G is the gravitational constant, m1 and m2 are the masses of the spheres, and r is the distance between them. Identify the coordinates of the spheres: 40 kg sphere at (鈭�1.0 m,鈭�1.0 m) and 60 kg sphere at (0 m,鈭�0.5 m).
02

- Calculate the Distances

Calculate the distance between the 20 kg sphere at the origin and each of the other spheres. Use the distance formula: distance formula: r = sqrt(((x2 鈥� x1)^2 + (y2 鈥� y1)^2) For the 40 kg sphere: r = sqrt((鈭�1.0 鈥� 0)^2 + (鈭�1.0 鈥� 0)^2)) = sqrt(1 + 1) = sqrt(2) 鈮� 1.41 meters. For the 60 kg sphere: r = sqrt((0 鈥� 0)^2 + (鈭�0.5 鈥� 0)^2)) = sqrt(0.25) = 0.5 meters.
03

- Calculate the Gravitational Forces

Calculate the gravitational forces using Newton鈥檚 Law of Gravitation. For the 40 kg sphere: F_40 = G frac((20)(40))((1.41)^2 = G (800/1.9881) = 400 G. For the 60 kg sphere: F_60 = G (60)(20) /(0.5) ^2) = G2400.
04

- Determine the Components

Determine the x and y components of each force. For F_40: F_x40 = (鈭�400 G (cos 45 )) = (鈭�400 G/sqrt 2) F_y40 = (鈭�400 G (sin 45 )) = (鈭�400 G/sqrt 2) For F_60: F_x60 = (0). F_y60 = 鈭�2400 G.
05

- Sum of Forces Components

Add the x and y components to find the resultant force components. F_x_total = F_x40 + F_x60 = 鈭�400 G/sqrt2 + 0 = 鈭�283 G. F_y_total = F_y40 + F_y60 = 鈭�400 G/sqrt2 鈭� 2400 G = 鈭� 282.8 G 鈭� 2400 G 鈮� 鈭�2682.8 G.
06

- Determine the Magnitude

Calculate the magnitude of the resultant force using the Pythagorean theorem. F_total = sqrt( F_x_total^2+F_y_total^2)) = sqrt(((鈭�283 G)^2) + (鈭�2683 G)^2)) = sqrt(8.9 x 10^5 G^2 + 7. 2 X 10^5 G^2) =%) 2625 G.

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Key Concepts

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

Newton's Law of Gravitation
Newton's Law of Gravitation is fundamental to understanding how objects in space interact through the force of gravity. It states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The formula used to calculate this gravitational force is \(F = G \frac{m_1 m_2}{r^2}\)where:
  • \(F\) is the gravitational force between the masses
  • \(G\) is the gravitational constant, approximately equal to \(6.674 \times 10^{-11} N (m/kg)^2\)
  • \(m_1\) and \(m_2\) are the masses of the two objects
  • \(r\) is the distance between the centers of the two masses
In this exercise, we used this law to find the gravitational forces exerted on a 20 kg sphere by two other spheres of masses 40 kg and 60 kg. The calculations were simplified by clearly identifying the positions and masses of the spheres.
Distance Formula
The distance formula is essential for calculating the distance between two points in a plane. This formula is derived from the Pythagorean theorem and is given by: \(r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)where:
  • \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points
  • \(r\) is the distance between the points
In our exercise, we calculated the distances between a 20 kg sphere at the origin and two other spheres at coordinates (鈭�1.0 m, 鈭�1.0 m) and (0 m, 鈭�0.5 m). For the 40 kg sphere, the distance was approximately 1.41 meters, and for the 60 kg sphere, it was 0.5 meters.
Force Components
After calculating the gravitational forces between the spheres, it's crucial to break these forces into their x and y components. Decomposing forces helps in understanding the direction and resultant magnitude of the forces. For this, we use trigonometric functions: \(F_x = F \cos(\theta)\)\(F_y = F \sin(\theta)\)where:
  • \(F\) is the total force exerted by the sphere
  • \(\theta\) is the angle of the force relative to the axis
  • \(F_x\) and \(F_y\) are the components of the force in the x and y directions, respectively
By applying these formulas, we calculated the x and y components of the forces exerted by the 40 kg and 60 kg spheres on a 20 kg sphere at the origin. Summing these components gave us the overall force's magnitude and direction acting on the 20 kg sphere.

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Most popular questions from this chapter

A solid sphere of uniform density has a mass of \(1.0 \times 10^{4} \mathrm{~kg}\) and a radius of \(1.0 \mathrm{~m}\). What is the magnitude of the gravitational force due to the sphere on a particle of mass \(m\) located at a distance of (a) \(1.5 \mathrm{~m}\) and (b) \(0.50 \mathrm{~m}\) from the center of the sphere? (c) Write a general expression for the magnitude of the gravitational force on the particle at a distance \(r \leq 1.0 \mathrm{~m}\) from the center of the sphere.

(A) What is the escape speed on a spherical asteroid whose radius is \(500 \mathrm{~km}\) and whose gravitational acceleration at the surface is \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of \(1000 \mathrm{~m} / \mathrm{s} ?(\mathrm{c})\) With what speed will an object hit the asteroid if it is dropped from \(1000 \mathrm{~km}\) above the surface?

Some believe that the positions of the planets at the time of birth influence the newborn. Others deride this belief and claim that the gravitational force exerted on a baby by the obstetrician is greater than that exerted by the planets. To check this claim, calculate and compare the magnitude of the gravitational force exerted on a \(3 \mathrm{~kg}\) baby (a) by a \(70 \mathrm{~kg}\) obstetrician who is \(1 \mathrm{~m}\) away and roughly approximated as a point mass, (b) by the massive planet Jupiter \(\left(m=2 \times 10^{27} \mathrm{~kg}\right)\) at its closest approach to Earth \(\left(=6 \times 10^{11} \mathrm{~m}\right)\), and \((\mathrm{c})\) by Jupiter at its greatest distance from Earth \(\left(=9 \times 10^{11} \mathrm{~m}\right)\). (d) Is the claim correct?

Black Hole The radius \(R_{h}\) of a black hole is the radius of a mathematical sphere, called the event horizon, that is centered on the black hole. Information from events inside the event horizon cannot reach the outside world. According to Einstein's general theory of relativity, \(R_{h}=2 G M / c^{2}\), where \(M\) is the mass of the black hole and \(c\) is the speed of light. Suppose that you wish to study black holes near them, at a radial distance of \(50 R_{k} .\) However, you do not want the difference in gravitational acceleration between your feet and your head to exceed \(10 \mathrm{~m} / \mathrm{s}^{2}\) when you are feet down (or head down) toward the black hole. (a) As a multiple of our sun's mass, what is the limit to the mass of the black hole you can tolerate at the given radial distance? (You need to estimate your height.) (b) Is the limit an upper limit (you can tolerate smaller masses) or a lower limit (you can tolerate larger masses)?

Zero, a hypothetical planet, has a mass of \(5.0 \times 10^{23} \mathrm{~kg}\), a radius of \(3.0 \times 10^{6} \mathrm{~m}\), and no atmosphere. A \(10 \mathrm{~kg}\) space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of \(5.0 \times 10^{7} \mathrm{~J}\), what will be its kinetic energy when it is \(4.0 \times 10^{6} \mathrm{~m}\) from the center of Zero? (b) If the probe is to achieve a maximum distance of \(8.0 \times 10^{6} \mathrm{~m}\) from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?
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