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Lunna
1 month ago
14

In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric

al, in this problem we'll model it as a sphere with a mass of 1.0 × 1013 kg and a radius of 1.6 km. If a rock were dropped from a height of 1.0 m above the comet's surface, how long would it take to hit the surface?
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
8 0

To tackle this issue, we will utilize concepts related to gravity based on Newtonian definitions. To find this value, we'll apply linear motion kinematic equations to determine the required time. Our parameters include:

Comet mass M = 1.0*10^{13} kg

Radius r = 1.6km = 1600 m

The rock is released from a height 'h' of 1 m above the surface.

The relationship for gravity's acceleration concerning a body with mass 'm' and radius 'r' is described by:

g = \frac{GM}{R^2}

Where G represents the gravitational constant and M denotes the mass of the planet.

g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}

g = 2.607*10^{-4} m/s^2

Now, let’s compute the time value.

h = \frac{1}{2} gt^2

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}

t = 87.58s

Ultimately, the time for the rock to hit the surface is t = 87.58s.

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