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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