Chapter 12: Problem 27
After a spacewalk, a 1.00 -kg tool is left \(50.0 \mathrm{~m}\) from the center of gravity of a 20.0 -metric ton space station, orbiting along with it. How much closer to the space station will the tool drift in an hour due to the gravitational attraction of the space station?
Short Answer
Expert verified
Answer: The tool drifts approximately 17.3 μm closer to the space station in an hour due to the gravitational attraction between them.
Step by step solution
01
Identify the given values
We are given:
- Mass of the tool, \(m_1 = 1.00 \hspace{2mm} kg\)
- Mass of the space station, \(m_2 = 20.0 \hspace{2mm} metric \hspace{2mm} tons = 20,000 \hspace{2mm} kg\)
- Initial distance between the center of mass of the tool and space station, \(r = 50.0 \hspace{2mm} m\)
- Time elapsed, \(t = 1 \hspace{2mm} hour = 3600 \hspace{2mm} seconds\)
02
Write down the Law of Universal Gravitation formula
The gravitational force between two objects with masses \(m_1\) and \(m_2\), separated by a distance \(r\), is given by the formula:
\(F = G\frac{m_1 \cdot m_2}{r^2}\)
where \(G\) is the gravitational constant, \(G = 6.674 \times 10^{-11} \hspace{2mm} N\cdot m^2/kg^2\).
03
Calculate the gravitational force between the tool and the space station
Using the given values, we can calculate the gravitational force between the tool and the space station:
\(F = (6.674 \times 10^{-11})\frac{1.00 \cdot 20,000}{50.0^2}\)
\(F = 2.6696 \times 10^{-9} \hspace{2mm} N\)
04
Calculate the acceleration of the tool towards the space station
Using Newton's second law of motion, \(F = m \cdot a\), we can find the acceleration of the tool towards the space station:
\(a = \frac{F}{m_1}\)
\(a = \frac{2.6696 \times 10^{-9}}{1.00}\)
\(a = 2.6696 \times 10^{-9} \hspace{2mm} m/s^2\)
05
Calculate the distance the tool drifts towards the space station in an hour
Using the equation of motion, \(d = \frac{1}{2} a \cdot t^2\), we can calculate the distance the tool drifts towards the space station in an hour:
\(d = \frac{1}{2} (2.6696 \times 10^{-9})(3600)^2\)
\(d \approx 17.3 \times 10^{-6} \hspace{2mm} m\)
The tool drifts approximately \(17.3 \mathrm{\mu m}\) closer to the space station in an hour due to the gravitational attraction between them.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Law of Universal Gravitation
Understanding the Law of Universal Gravitation is essential in predicting how celestial objects interact with each other due to the force of gravity. According to this law, formulated by Sir Isaac Newton, 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.
This concept is expressed mathematically as: \[\begin{equation}F = G \frac{m_1 \cdot m_2}{r^2}\end{equation}\]where:
This concept is expressed mathematically as: \[\begin{equation}F = G \frac{m_1 \cdot m_2}{r^2}\end{equation}\]where:
- \( F \) is the gravitational force between the two masses;
- \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} N \cdot m^2/kg^2 \));
- \( m_1 \) and \( m_2 \) are the masses of the two objects;
- \( r \) is the distance between the centers of the masses.
Newton's Second Law of Motion
Newton's second law of motion provides the relationship between an object's mass, the net force acting upon it, and its acceleration. The net force applied to an object causes it to accelerate in the direction of the force. This is expressed by the equation:\[\begin{equation}F = m \cdot a\end{equation}\]where:
- \( F \) stands for the force exerted on the object;
- \( m \) represents the mass of the object;
- \( a \) is the acceleration experienced by the object.
Equation of Motion
The equation of motion is a mathematical expression used to calculate the displacement of an object when its acceleration (a) and the time of motion (t) are known. The simplest form, assuming the initial velocity is zero and the acceleration is constant, is given by:\[\begin{equation}d = \frac{1}{2} a \cdot t^2\end{equation}\]where:
- \( d \) is the displacement;
- \( a \) is the constant acceleration;
- \( t \) is the time elapsed.