An astronaunt lands on the dwarf planet Eris and decides to perform experiments.

Question

An astronaunt lands on the dwarf planet Eris and decides to perform experiments. One of these experiments is to measure the acceleration due to gravity. He finds that the acceleration due to gravity is constant at 0.8 m/s2. Then he finds a deep crater and wants to know its depth so he fires a rope to the center of the crater. He then slides down the rope without friction and finds that his velocity at the bottom is 63.79 ± 0.27 m/s. Using this information and the fact that the astronaunt has a weight with his suit of 240 N on Mars, find the average depth of the crater.

Explanation / Answer

The acceleration due to gravity on the dwarf planet is g = 0.8 m/s2 Let h be the depth of the crate The speed of the astronaut at the bottom is v = 63.79±0.27 m/s The weight of the astronaunt with the suit on Mars is W = 240 N where the acceleration due to gravity is 3.711 m/s2 The mass of the astronaunt with the suit is M = 64.67 kg As the astronaunt slides down without friction, the case resembles a free fall So the speed at the bottom is given by v = 2gh h = v2/2g For the uncertainities, we have hmax = vmax2/2g hmax = (64.06 m/s)2/2(0.8 m/s2) hmax = 2564.8 m hmin = vmin2/2g hmin = (63.52 m/s)2/2(0.8 m/s2) hmin = 2521.7 m The average depth of the crate is h = (2564.8 m + 2521.7 m)/2 = 2543.27 m Note: the depth of the crate with uncertainity will be h =   (2564.8 m + 2521.7 m)/2 ± (2564.8 m - 2521.7 m)/2 h = 2543.27 ± 21.53 m The mass of the astronaunt with the suit is M = 64.67 kg As the astronaunt slides down without friction, the case resembles a free fall So the speed at the bottom is given by v = 2gh h = v2/2g For the uncertainities, we have hmax = vmax2/2g hmax = (64.06 m/s)2/2(0.8 m/s2) hmax = 2564.8 m hmin = vmin2/2g hmin = (63.52 m/s)2/2(0.8 m/s2) hmin = 2521.7 m The average depth of the crate is h = (2564.8 m + 2521.7 m)/2 = 2543.27 m Note: the depth of the crate with uncertainity will be h =   (2564.8 m + 2521.7 m)/2 ± (2564.8 m - 2521.7 m)/2 h = 2543.27 ± 21.53 m hmin = vmin2/2g hmin = (63.52 m/s)2/2(0.8 m/s2) hmin = 2521.7 m The average depth of the crate is h = (2564.8 m + 2521.7 m)/2 = 2543.27 m Note: the depth of the crate with uncertainity will be h =   (2564.8 m + 2521.7 m)/2 ± (2564.8 m - 2521.7 m)/2 h = 2543.27 ± 21.53 m

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