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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 \(1.13 \times 10^{-16}\,\text{ly/yr}^2\).

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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.
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:
  • 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: \(\text{ly/yr}^2\).
Addressing such problems demands attention to detail and understandings, like how speed, acceleration, and distance interact in both terrestrial and celestial contexts.
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:
  • 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.
The application of significant figures not only helps maintain uniformity in scientific work but also ensures results are communicated with the appropriate precision.

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