If a planet has 3 times the radius of the Earth,but has the same density as the Earth,what is the gravitational acceleration at the surface of the planet? (g = 9.80 m/s²) A)29.4 m/s² B)88.2 m/s² C)265 m/s² D)3.27 m/s²

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Quizplus · Accepted Answer

The gravitational acceleration at the surface of a planet can be calculated using the formula $g' = g \times \frac{M'}{M} \times (\frac{R}{R'})^2$, where $g$ is the Earth's gravitational acceleration, $M'$ and $M$ are the masses of the planet and Earth respectively, and $R'$ and $R$ are the radii of the planet and Earth respectively. Given that the planet has 3 times the radius of Earth and the same density, its volume, and thus its mass, will be $3^3 = 27$ times that of Earth. Substituting these values into the formula gives $g' = 9.8 \times 27 \times (\frac{1}{3})^2 = 29.4 \, m/s^2$.

If a Planet Has 3 Times the Radius of the Earth,but