Chapter 35: Problem 22
Find the value of \(g\), the gravitational acceleration at Earth's surface, in light-years per year, to three significant figures.
Short Answer
Expert verified
Answer: 1.13 x 10^-16 ly/yr²
Step by step solution
01
Determining Known Values and Conversion Factors
The known values required are:
- \(g\) (Earth's gravitational acceleration) in meters per second squared (m/s^2): \(g = 9.81\,\text{m/s}^2\).
- \(c\) (speed of light) in meters per second (m/s): \(c = 3.00\times10^8\,\text{m/s}\).
- \(t\) (number of seconds in a year): \(t = 3.15\times10^7\,\text{s}\).
With these known values, we can come up with the necessary conversion factors:
- Meters to light-years: \(1\,\text{light-year} = \frac{c \times t}{1\,\text{year}}\).
- Seconds to years: \(1\,\text{year} = 3.15 \times 10^7\,\text{s}\).
Step 2:
02
Converting Earth's Gravitational Acceleration to Light-Years Per Year
We will first find the conversion factor for meters to light-years:
\(1\,\text{light-year} = \frac{c \times t}{1\,\text{year}} = \frac{3.00\times10^8\,\text{m/s} \times 3.15\times10^7\,\text{s}}{1\,\text{year}} = 9.46\times10^{15}\,\text{m}\).
Now, we can convert the value of \(g\) from meters per second squared to light-years per year squared:
\(g_{\text{ly/yr}^2} = \left(\frac{9.81\,\text{m/s}^2}{9.46\times10^{15}\,\text{m/ly}}\right)\times\left(\frac{1\,\text{year}}{3.15\times10^7\,\text{s}}\right)^2\)
Step 3:
03
Rounding the Result to Three Significant Figures
Evaluating the previous expression, we get:
\(g_{\text{ly/yr}^2} \approx 1.1251 \times 10^{-16}\,\text{ly/yr}^2\)
To round the result of \(g\) to three significant figures, we get:
\(g_{\text{ly/yr}^2} \approx 1.13 \times 10^{-16}\,\text{ly/yr}^2\)
So, the value of Earth's gravitational acceleration in light-years per year squared, to three significant figures, is \(1.13 \times 10^{-16}\,\text{ly/yr}^2\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Light-year Conversion
To understand light-year conversion, it’s crucial to recognize that a light-year is not a unit of time, but a unit of distance. It is defined as the total distance light travels in one year through the vacuum of space. Given that light speed is approximately 300,000 kilometers per second, a single light-year is a vast expanse.
To convert units like meters to light-years, we rely on fixed physical values. One such value is the speed of light in meters per second (\(c\)). Another important value is the total seconds in a year, as the journey of light is time-dependent. By multiplying the speed of light by seconds per year, we derive how far light travels in one year, which equals one light-year.
The calculations involved in converting gravitational acceleration, like Earth's gravity (\(g\)), to light-years per year squared, require substituting and manipulating these constants. It underscores the vastness of the universe in comprehensible earthly terms.
To convert units like meters to light-years, we rely on fixed physical values. One such value is the speed of light in meters per second (\(c\)). Another important value is the total seconds in a year, as the journey of light is time-dependent. By multiplying the speed of light by seconds per year, we derive how far light travels in one year, which equals one light-year.
The calculations involved in converting gravitational acceleration, like Earth's gravity (\(g\)), to light-years per year squared, require substituting and manipulating these constants. It underscores the vastness of the universe in comprehensible earthly terms.
Physics Problems
Physics problems often involve a juxtaposition of various units and constants, bringing real-world scenarios into calculable expressions. In our case, we're integrating a terrestrial measurement, Earth's gravitational acceleration (\(9.81 \, \text{m/s}^2\)), with astronomical distances. Tackling such problems enhances our ability to grasp the interconnectedness of different physical domains.
In steps, we break down the conversion by:
In steps, we break down the conversion by:
- Identifying constants: gravitational constants and speed of light.
- Implementing dimensional analysis: transforming meters to light-years, and seconds to years.
- Employing algebraic manipulation to recombine these into the desired units: \(\text{ly/yr}^2\).
Significant Figures
The concept of significant figures is crucial for precision in scientific calculations. Significant figures help in communicating the accuracy of a measurement based on the precision of the measuring tool.
In our gravitational acceleration example, the final estimation is presented as \(1.13 \times 10^{-16} \, \text{ly/yr}^2\). This ensures that the expression reflects the three significant figures required.
To round to significant figures:
In our gravitational acceleration example, the final estimation is presented as \(1.13 \times 10^{-16} \, \text{ly/yr}^2\). This ensures that the expression reflects the three significant figures required.
To round to significant figures:
- Identify the digits in a number that are meaningful in terms of accuracy.
- Round the number so that only the specified significant figures remain.
- Ensure the rounding process respects the accuracy implied by initial measurements.