Question

In 2014, the Rosetta space probe reached the comet Churyumov-Gerasimenko. Although the comet's core is actually far from spherical, in this problem we'll model it as a sphere with a mass of 1.0 x 10₁₃ kg and a radius pf 1.6 km Part A. If a rock were dropped from a height of 0.80 m above the comet's surface, how long would it take to hit the surface?

Answer 1

Physics principles and kinematic equations are used to determine the time it takes for a rock to fall on the comet's surface, considering the comet's much weaker gravitational pull compared to Earth.

Step-by-step explanation:

The student's question about the time it would take for a rock to fall on Comet Churyumov-Gerasimenko requires an understanding of gravitational acceleration and kinematics, which is a topic covered in physics. To solve this, we must consider the comet's gravitational field as it would be much weaker than Earth's due to its smaller mass and size.

To calculate the time of fall, we use the formula derived from the kinematic equations:

1. d = (1/2)gt^2,
2. Where: d is the distance fallen, g is the gravitational acceleration, and t is the time in seconds.

To find the acceleration, we use Newton's law of universal gravitation:

1. g = G * (M/R^2),
2. Where: G is the gravitational constant, M is the mass of the comet, and R is the radius of the comet.

Given Rosetta's mission data, the comet's mass (M) is 1.0 x 10^13 kg and the radius (R) is 1.6 x 10^3 meters. Plugging these into the formula for g, and then g into the kinematic formula, we can solve for t.

The reduced gravity on the comet would result in a much longer fall time for the rock compared to the same drop on Earth, even for the relatively short distance of 0.80 meters.