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Floating in space far from anything else are two spherical asteroids, one having a mass of \(20 \times 10^{10} \mathrm{~kg}\) and the other a mass of \(40 \times 10^{10} \mathrm{~kg}\). Compute the force of attraction on each one due to gravity when their center-to-center separation is \(10 \times 10^{6} \mathrm{~m}\).

Short Answer

Expert verified
The gravitational force between the asteroids is approximately \( 5.3392 \times 10^{-2} \, \text{N} \).

Step by step solution

01

Identify the Given Variables

We have two spherical asteroids with masses \(m_1 = 20 \times 10^{10} \, \text{kg}\) and \(m_2 = 40 \times 10^{10} \, \text{kg}\). The distance between the centers of these asteroids is \(r = 10 \times 10^{6} \, \text{m}\).
02

Recall the Formula for Gravitational Force

The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by Newton's Law of Universal Gravitation: \[ F = G \frac{m_1 m_2}{r^2} \] where \( G = 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2\) is the gravitational constant.
03

Substitute the Given Values into the Formula

Substitute \( m_1 = 20 \times 10^{10} \, \text{kg} \), \( m_2 = 40 \times 10^{10} \, \text{kg} \), \( r = 10 \times 10^{6} \, \text{m} \), and \( G = 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2 \) into the formula: \[F = 6.674 \times 10^{-11} \frac{(20 \times 10^{10})(40 \times 10^{10})}{(10 \times 10^{6})^2}\]
04

Calculate the Gravitational Force

Perform the calculation step-by-step:- Calculate the product of the masses: \ \( m_1 m_2 = (20 \times 10^{10})(40 \times 10^{10}) = 800 \times 10^{20} \, \text{kg}^2\)- Calculate the square of the distance: \ \( r^2 = (10 \times 10^{6})^2 = 100 \times 10^{12} \, \text{m}^2\)- Plug these into the gravitational formula: \[F = 6.674 \times 10^{-11} \frac{800 \times 10^{20}}{100 \times 10^{12}}\]
05

Simplify and Solve for the Force

Continue simplifying the expression: - Simplify the fraction: \ \( \frac{800 \times 10^{20}}{100 \times 10^{12}} = 8 \times 10^{8} \)- Multiply by \( G \): \[F = 6.674 \times 10^{-11} \times 8 \times 10^{8} = 5.3392 \times 10^{-2} \, \text{N}\] Thus, the force of attraction between the asteroids is \( 5.3392 \times 10^{-2} \, \text{N} \).

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Key Concepts

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

Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation is a fundamental principle in physics that describes the attractive force between two masses. This law states that every mass attracts every other mass in the universe, and the force between the two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This is expressed mathematically as:
  • \[ F = G \frac{m_1 m_2}{r^2} \]
Here, \( F \) is the gravitational force, \( m_1 \) and \( m_2 \) are the masses of the two objects, \( r \) is the distance between the centers of the masses, and \( G \) is the gravitational constant. This law is crucial for understanding how objects attract each other in space, such as planets, stars, and in our exercise, asteroids. By applying this formula, we can calculate the gravitational pull between any two celestial bodies.
Gravitational Constant
The gravitational constant, represented by \( G \), is a key value in the formula for calculating gravitational force. It measures the strength of gravity in the universal framework and is a fixed number in the equation.
  • The standard value of \( G \) is \( 6.674 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2 \).
This constant ensures that the units in the formula align correctly, providing a resulting force in newtons (N). Understanding \( G \) is important because it allows us to calculate the gravitational interactions not just on Earth but throughout the cosmos. It's a very small number, indicating that gravity is a weak force compared to other fundamental forces like electromagnetism, yet it plays a significant role in the universe due to the large masses involved in celestial bodies.
Spherical Asteroids
In our exercise, the gravitational attraction is calculated between two spherical asteroids. The assumption of spherical shape simplifies calculations, as it allows us to assume that the entire mass of the asteroid can be considered to be concentrated at its center of mass. This simplification is crucial in applying Newton's Law of Universal Gravitation straightforwardly.
  • The shape simplifies mathematical modeling, making calculations feasible with basic geometry concepts.
When dealing with real celestial objects, they are often approximated as spheres to use this law effectively. This approach works particularly well for large distances, such as the separation between our asteroids, where smaller irregularities of shape become negligible. Understanding this concept helps to model interactions in space without requiring complex geometry to account for the actual shape of the bodies involved.
Step-by-Step Calculation
Performing a step-by-step calculation helps in understanding how gravitational force is determined. This structured approach allows us to carefully follow each part of the computation:
  • First, identify the given masses and distance.
  • Next, apply the formula \[ F = G \frac{m_1 m_2}{r^2} \], substituting these values.
In our problem with the asteroids, we calculated:
  • The product of their masses \( m_1 m_2 = 800 \times 10^{20} \, \text{kg}^2 \).
  • The square of the distance \( r^2 = 100 \times 10^{12} \, \text{m}^2 \).
Then, we plugged these into the formula and simplified:
  • \[ F = 6.674 \times 10^{-11} \frac{800 \times 10^{20}}{100 \times 10^{12}} = 5.3392 \times 10^{-2} \, \text{N} \]
This methodical process clarifies how individual components contribute, ensuring the final result is accurate. The clarity of following a structured method enhances comprehension, vital for accurately applying similar calculations in future tasks.

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Most popular questions from this chapter

Once ignited, a small rocket motor on a spacecraft exerts a constant force of \(10 \mathrm{~N}\) for \(7.80 \mathrm{~s}\). During the burn, the rocket causes the \(100-\mathrm{kg}\) craft to accelerate uniformly. Determine that acceleration.

The coefficient of static friction between a box and the flat bed of a truck is \(0.60\). What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{\mathrm{f}}=\mu_{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck: otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{\mathrm{f}}=m a_{x}\) However, if the box is on the verge of slipping, \(F_{\mathrm{f}}=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.

Two forces, \(80 \mathrm{~N}\) and \(100 \mathrm{~N}\), acting at an angle of \(60^{\circ}\) with each other, pull on an object. ( \(a\) ) What single force would replace the two forces? (b) What single force (called the equilibrant) would balance the two forces? Solve algebraically.

An elevator starts from rest with a constant upward acceleration. It moves \(2.0 \mathrm{~m}\) in the first \(0.60 \mathrm{~s}\). A passenger in the elevator is holding a \(3.0-\mathrm{kg}\) package by a vertical string. What is the tension in the string during the accelerating process?

An inclined plane making an angle of \(25^{\circ}\) with the horizontal has a pulley at its top. A \(30-\mathrm{kg}\) block on the plane is connected to a freely hanging 20 -kg block by means of a cord passing over the pulley. Compute the distance the \(20-\mathrm{kg}\) block will fall in \(2.0 \mathrm{~s}\) starting from rest. Neglect friction.

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