Question

Find the value of \(g\), the gravitational acceleration at Earth's surface, in light-years per year per year, to three significant figures.

Short answer

Answer: 1.03 x 10⁻¹⁶ light-years/year².

Step-by-step solution

Write down the known value of gravitational acceleration in standard units.

The standard value of gravitational acceleration (g) at Earth's surface is approximately 9.81 m/s².

Convert meters per second squared to light-years per year squared.

First, we need to convert meters to light-years. There are 9.461 x 10¹² kilometers in one light-year. Since there are 1000 meters in a kilometer, there are 9.461 x 10¹² x 1000 meters in one light-year which is equivalent to 9.461 x 10¹⁵ meters. Now, we must convert seconds to years. There are 60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and 365.25 days in a year (accounting for leap years), so there are 60 x 60 x 24 x 365.25 seconds in a year, which is equivalent to 3.154 x 10⁷ seconds. Next, we will convert 9.81 m/s² to light-years per year squared: (9.81 m/s²) * (1 light-year / 9.461 x 10¹⁵ m) * (3.154 x 10⁷ s / 1 year)² = 1.029 x 10⁻¹⁶ light-years/year².

Round the result to three significant figures.

Our result is 1.029 x 10⁻¹⁶ light-years/year². To round this to three significant figures, we get: 1.03 x 10⁻¹⁶ light-years/year². The value of g, the gravitational acceleration at Earth's surface, is approximately 1.03 x 10⁻¹⁶ light-years per year per year.