An object drops a distance h in a time of 6.3 s on the surface of the earth (neglecting air effects).How long would it take the same object to drop the same distance on the surface of spherical asteroid X having a mass of 1.1 × 10²² kg and a radius of 4.0 × 10⁵ m? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg,R_earth = 6.38 × 10⁶ m) A)9.2 s B)6.3 s C)5.2 s D)12 s E)3.3 s

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Quizplus · Accepted Answer

The time it takes for an object to drop a certain distance is related to the acceleration due to gravity on the surface of the celestial body. The acceleration due to gravity, $g$, can be calculated using the formula $g = \frac{G \cdot M}{R^2}$, where $G$ is the gravitational constant, $M$ is the mass of the celestial body, and $R$ is its radius.For Earth, $g_{\text{earth}} = \frac{6.67 \times 10^{-11} \cdot 5.97 \times 10^{24}}{(6.38 \times 10^6)^2} \approx 9.8 \, \text{m/s}^2$.For asteroid X, $g_{\text{X}} = \frac{6.67 \times 10^{-11} \cdot 1.1 \times 10^{22}}{(4.0 \times 10^5)^2}$.The time to fall a distance $h$ is given by $t = \sqrt{\frac{2h}{g}}$. Since $h$ is constant, the ratio of times on Earth and asteroid X is proportional to the square root of the inverse ratio of their gravities, i.e., $t_{\text{X}} = t_{\text{earth}} \cdot \sqrt{\frac{g_{\text{earth}}}{g_{\text{X}}}}$.Calculating $g_{\text{X}}$ and then $t_{\text{X}}$ with the given values, you find that the time on asteroid X is longer due to its smaller mass and radius, which results in a lower gravitational acceleration compared to Earth. The exact calculation would show that the correct answer is 9.2 seconds, reflecting the reduced gravity's effect on the object's fall time.

An Object Drops a Distance H in a Time of 6.3