Question

At a certain instant, the earth, the moon, and a stationary 1250-kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 \(\times\) 10\(^5\) 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? Ignore any gravitational effects due to the other planets or the sun.

Short answer

(a) Net force: 0.130 N at an angle 1.15° from the line connecting Earth and spacecraft. (b) Work needed: 8.34 × 10⁹ J.

Step-by-step solution

Identify the Gravitational Force Formula

The gravitational force between two objects with masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by Newton's law of gravitation: \[ F = \frac{G m_1 m_2}{r^2} \] where \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{m}^3 \cdot \text{kg}^{-1} \cdot \text{s}^{-2} \).

Convert the Distance to Meters

The problem gives the side length of the triangle as \( 3.84 \times 10^5 \) km. Convert this to meters to be compatible with our unit system: \[ 3.84 \times 10^5 \, \text{km} = 3.84 \times 10^8 \, \text{m} \]

Calculate the Gravitational Force from the Earth

The mass of Earth \( m_E \) is \( 5.972 \times 10^{24} \) kg. Using the formula from Step 1: \[ F_{E} = \frac{6.674 \times 10^{-11} \times 1250 \times 5.972 \times 10^{24}}{(3.84 \times 10^8)^2} \approx 0.134 \, \text{N} \]

Calculate the Gravitational Force from the Moon

The mass of the Moon \( m_M \) is \( 7.348 \times 10^{22} \) kg. Using the same formula: \[ F_{M} = \frac{6.674 \times 10^{-11} \times 1250 \times 7.348 \times 10^{22}}{(3.84 \times 10^8)^2} \approx 0.0027 \, \text{N} \]

Calculate the Net Gravitational Force

Since the forces from the Earth and the Moon are at 60° to each other (the angles of an equilateral triangle), use vector addition. The net force can be calculated using: \[ F_{net} = \sqrt{F_{E}^2 + F_{M}^2 + 2 F_{E} F_{M} \cos(60°)} \approx 0.130 \, \text{N} \]

Find the Direction of the Net Force

The direction of \( F_{net} \) can be found using: \[ \theta = \tan^{-1}\left(\frac{F_{M} \sin(60°)}{F_{E} + F_{M} \cos(60°)}\right) \] \[ \theta \approx 1.15° \] from the line connecting the earth and spacecraft.

Calculate the Work to Move the Spacecraft

The minimum work is equal to the gravitational potential energy of the system. Since the spacecraft is to be moved far from both Earth and Moon: \[ W = -G \left(\frac{m_E m_{S}}{d} + \frac{m_M m_{S}}{d}\right) \] \[ W = -\left(\frac{6.674 \times 10^{-11} \times 1250 \times 5.972 \times 10^{24}}{3.84 \times 10^8} + \frac{6.674 \times 10^{-11} \times 1250 \times 7.348 \times 10^{22}}{3.84 \times 10^8}\right) \] \[ W \approx 8.34 \times 10^{9} \, \text{J} \]

Key concepts

Newton's law of gravitation

Newton's law of gravitation is a cornerstone of classical physics. It explains how two bodies attract each other due to their masses. The force of attraction is given by the formula:

• \[ F = \frac{G m_1 m_2}{r^2} \]
• Here, \( F \) is the gravitational force, \( G \) is the gravitational constant \((6.674 \times 10^{-11} \, \text{m}^3 \cdot \text{kg}^{-1} \cdot \text{s}^{-2})\), \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between the centers of the two masses.

This law shows that gravitational force decreases as the distance between two objects increases. Also, the force is directly proportional to the product of their masses. In the exercise, the Earth and Moon separately attract the spacecraft. Calculating these forces allows us to understand their combined effects on the spacecraft's motion.

work-energy principle

The work-energy principle helps us to find the work needed to move an object over a distance under the influence of forces like gravity. Work in physics is calculated as the force applied over a distance. For gravity, the concept extends to gravitational potential energy.

• The work-energy principle states that work done on an object is equal to the change in its kinetic energy, expressed as:
• \[ W = \Delta KE = KE_{final} - KE_{initial} \]
• However, in gravitational contexts, we often refer to potential energy changes.
• The amount of work required to move the spacecraft far from Earth and Moon involves overcoming the gravitational potential energy stored due to its position. This work can be found by summing the potential energies due to Earth and the Moon.

Addressing this principle helps us comprehend the energy requirements for moving objects in gravitational fields.

vector addition in physics

In physics, vector addition is crucial for calculating net forces that involve multiple vectors acting at angles to one another. Gravitational forces exerted on an object are vectors, meaning they have both magnitude and direction.

• To find the net gravitational force on the spacecraft, we need to consider both the Earth's and the Moon's forces. As described in the exercise, these forces form an angle of 60° because of the equilateral triangle configuration.
• The vector addition formula used is given by:
• \[ F_{net} = \sqrt{F_{E}^2 + F_{M}^2 + 2 F_{E} F_{M} \cos(60°)} \]
• Through this, we calculate both the magnitude and direction of the resulting net force.

This concept allows accurate determination of how forces combine when they don't align perfectly, as is often the case in real-world scenarios.

gravitational potential energy

Gravitational potential energy pertains to the energy stored due to the position of an object in a gravitational field. It reflects the work required to move an object in such fields.

• The formula for gravitational potential energy is:
• \[ U = -G \frac{m_1 m_2}{r} \]
• Here, \( U \) is the potential energy, and the negative sign indicates that work is needed to move the object to an infinite distance (where potential energy is zero).
• For the spacecraft in the exercise, the potential energy due to both Earth and the Moon is considered to calculate the minimum work required to move the spacecraft away.

Understanding this concept helps us appreciate energy changes connected to object movements under gravity, crucial for tasks such as spacecraft launches and satellite deployments.