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At a certain instant, the earth, the moon, and a stationary \(1250-\mathrm{kg}\) spacecraft lie at the vertices of an equilateral triangle whose sides are \(3.84 \times 10^{5} \mathrm{km}\) in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun.

Short Answer

Expert verified
The net gravitational force on the spacecraft is approximately \(1.45 \times 10^3 \, \text{N}\), directed at an angle of \(\theta\) from the Earth. The work required to move the spacecraft to infinity is approximately \(8.70 \times 10^9 \, \text{J}\).

Step by step solution

01

Define Constants and Variables

Begin by noting the given values and constants:- Mass of the spacecraft, \( m_s = 1250 \, \text{kg} \)- Mass of the Earth, \( M_e = 5.97 \times 10^{24} \, \text{kg} \)- Mass of the Moon, \( M_m = 7.35 \times 10^{22} \, \text{kg} \)- Distance between each body in the triangle, \( R = 3.84 \times 10^5 \, \text{km} = 3.84 \times 10^8 \, \text{m} \)- Gravitational constant, \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
02

Calculate Gravitational Force from Earth

The gravitational force exerted by the Earth on the spacecraft is given by: \[ F_{e} = \frac{G \cdot M_e \cdot m_s}{R^2} \]Plug in the known values:\[ F_{e} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 1250}{(3.84 \times 10^{8})^2} \approx 1.35 \times 10^3 \, \text{N} \]
03

Calculate Gravitational Force from Moon

The gravitational force exerted by the Moon on the spacecraft is given by:\[ F_{m} = \frac{G \cdot M_m \cdot m_s}{R^2} \]Plug in the known values:\[ F_{m} = \frac{6.674 \times 10^{-11} \times 7.35 \times 10^{22} \times 1250}{(3.84 \times 10^{8})^2} \approx 1.61 \times 10^2 \, \text{N} \]
04

Determine the Net Force Vector

Since the problem forms an equilateral triangle, the angle between the forces \( F_e \) and \( F_m \) is \( 60^\circ \). Use vector addition:\[ \vec{F}_{net} = \vec{F}_e + \vec{F}_m \]The magnitude of the net force is:\[ F_{net} = \sqrt{F_e^2 + F_m^2 + 2 \cdot F_e \cdot F_m \cdot \cos(60^\circ)} \]Plug in the values:\[ F_{net} = \sqrt{(1.35 \times 10^3)^2 + (1.61 \times 10^2)^2 + 2 \cdot 1.35 \times 10^3 \cdot 1.61 \times 10^2 \cdot 0.5} \approx 1.45 \times 10^3 \, \text{N} \]
05

Calculate the Angle of Direction

The angle \( \theta \) from the Earth-force vector can be found using:\[ \tan(\theta) = \frac{F_m \cdot \sin(60^\circ)}{F_e + F_m \cdot \cos(60^\circ)} \]\[ \theta = \tan^{-1}\left(\frac{1.61 \times 10^2 \times \sin(60^\circ)}{1.35 \times 10^3 + 1.61 \times 10^2 \times 0.5}\right) \]Calculate \( \theta \) to find the angle from the line joining Earth and the spacecraft.
06

Calculate Work to Move Spacecraft to Infinity

The minimum work done against gravitational forces to move the spacecraft to infinity is given by:\[ W = U = -\frac{G \cdot M_e \cdot m_s}{R} - \frac{G \cdot M_m \cdot m_s}{R} \]The total work:\[ W = G \cdot m_s \left(\frac{M_e + M_m}{R} \right) \]Substituting the values:\[ W \approx 6.674 \times 10^{-11} \cdot 1250 \cdot \left(\frac{5.97 \times 10^{24} + 7.35 \times 10^{22}}{3.84 \times 10^8}\right) \approx 8.70 \times 10^9 \, \text{J} \]

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Key Concepts

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

Vector Addition
In physics, especially when dealing with forces, vector addition is a crucial concept. Forces, like gravity, are vector quantities, meaning they have both magnitude and direction. When you have more than one force acting on an object, such as the spacecraft in the problem, you need to find the net force by adding these vector forces together.

For two forces acting at an angle, the total force, or net force, isn't just the arithmetic sum of the magnitudes of the individual forces due to their directional nature. You can use the triangle method or the parallelogram method to find the resultant vector.

In our example, we use the triangle method:
  • Construct a triangle with each force as a side.
  • The net force is the vector drawn from the starting point to the opposite end of the triangle.
  • Since our forces make an angle of 60 degrees in this equilateral triangle scenario, we use trigonometric functions and the law of cosines to find the resultant.
This shows that understanding vector addition is necessary for accurately describing the effects of forces acting on objects from different directions.
Work Against Gravity
Work against gravity is the amount of work needed to move a mass from one location to another against the force of gravity. In this context, it's about moving the spacecraft from its position in the gravitational fields of both Earth and the Moon to a point far away where these forces no longer exert influence.

The work done is equivalent to the gravitational potential energy, which can be calculated using: \[ W = U = -\frac{G \cdot M_e \cdot m_s}{R} - \frac{G \cdot M_m \cdot m_s}{R} \] This equation considers the potential energy due to Earth's and Moon's gravitational fields and subtracts these to reflect the removal from both fields. Gravitational potential energy is zero at infinity, making the total work equal to the initial gravitational potential energy at the spacecraft's starting point.
  • The higher the gravitational forces, the more work required to move the object.
  • This work is necessary to overcome the pull of gravity.
Understanding work against gravity is key to solving problems related to moving objects through gravitational fields.
Net Force Calculation
The net force on an object is the vector sum of all forces acting on that object. Calculating it requires both the magnitude and direction of each force—a challenge, especially when forces are not aligned in the same direction.

In our given problem, we have two forces—the gravitational pull from both Earth and the Moon—acting on the spacecraft. The calculation is as follows:
- Calculate the gravitational force from Earth using the formula: \[ F_{e} = \frac{G \cdot M_e \cdot m_s}{R^2} \]- Calculate the force from the Moon similarly.
Once you have both forces, use vector addition (considering they form an angle of 60 degrees in the triangle) to calculate the net force:
\[ F_{net} = \sqrt{F_e^2 + F_m^2 + 2 \cdot F_e \cdot F_m \cdot \cos(60^\circ)} \] Because the spacecraft forms part of an equilateral triangle with the Earth and Moon, the angle aids in simplifying the use of trigonometric identities. Clarity on both magnitude and direction of forces ensures the accuracy of the net force calculation.
Equilateral Triangle
An equilateral triangle is a triangle where all three sides are of equal length, and all angles are 60 degrees. This geometric property is highly useful in physics problems because it simplifies the use of trigonometric functions.

In this exercise, the Earth, Moon, and spacecraft form an equilateral triangle. This consistent geometry helps simplify the net force calculation and angle determination:
- The equal side lengths allow consistent calculations across the triangle. - The 60-degree angles mean that forces calculated using trigonometry will be consistent across similar problems, allowing straightforward use of the cosine rule for force computation.
The symmetry in an equilateral triangle typically leads to fewer variables and more straightforward calculations. This understanding aids in both setting up the problem correctly and ensuring that the mathematical approach remains consistent with the physical configuration.

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Most popular questions from this chapter

An experiment using the Cavendish balance to measure the gravitational constant \(G\) found that a uniform \(0.400-\mathrm{kg}\) sphere attracts another uniform \(0.00300-\mathrm{kg}\) sphere with a force of \(8.00 \times 10^{-10} \mathrm{N},\) when the distance between the centers of the spheres is 0.0100 \(\mathrm{m}\) . The acceleration due to gravity at the earth's surface is \(9.80 \mathrm{m} / \mathrm{s}^{2},\) and the radius of the earth is 6380 \(\mathrm{km}\) . Compute the mass of the earth from these data.

A uniform, solid, \(1000.0-\mathrm{kg}\) sphere has a radius of 5.00 \(\mathrm{m}\) . (a) Find the gravitational force this sphere exerts on a \(2.00-\mathrm{kg}\) point mass placed at the following distances from the center of the sphere: ( i) \(5.01 \mathrm{m},\) and (ii) 2.50 \(\mathrm{m} .\) (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 sppere. Include the region from \(r=0\) to \(r \rightarrow \infty .\)

A particle of mass 3\(m\) is located 1.00 \(\mathrm{m}\) from a particle of mass \(m .\) (a) Where should you put a third mass \(M\) so that the net gravitational force on \(M\) due to the two masses is exactly zero? (b) Is the equilibrium of \(M\) at this point stable or unstable (i) for points along the line connecting \(m\) and \(3 m,\) and (ii) for points along the line passing through \(M\) and perpendicular to the line connecting \(m\) and 3\(m ?\)

A thin, uniform rod has length \(L\) and mass \(M .\) A small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. 12.34\()\) . (a) Calculate the gravita- tional potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L\) . (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B. (b) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4 ). Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\)

(a) Calculate how much work is required to launch a spacecraft of mass \(m\) from the surface of the earth (mass \(m_{\mathrm{E}},\) radius \(R_{\mathrm{E}} )\) and place it in a circular low earth orbit - that is, an orbit whose altitude above the earth's surface is much less than \(R_{\mathrm{E}}\) . (As an example, the International Space Station is in low earth orbit at an altitude of about \(400 \mathrm{km},\) much less than \(R_{\mathrm{E}}=6380 \mathrm{km} .\) You can ignore the kimetic energy that the spacecraft has on the ground due to the earth's rotation. (b) Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. You can ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement: "In terms of energy, low earth orbit is halfway to the edge of the universe."

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