1. You are an astronaut (m = 85 kg) and travel to a planet that is the same mass
ID: 1690228 • Letter: 1
Question
1. You are an astronaut (m = 85 kg) and travel to a planet that is the same mass and size as Earth, but it has a rotational period of only 3 h. What is your apparent weight at the equator of this planet?( i have tried the other solutions posted and they don't work, please solve out problem)
2. Your apparent weight is the force you feel on the bottoms of your feet when you are standing. Due to the Earth's rotation, your apparent weight is slightly more when you are at the South Pole than when you are at the equator. What is the ratio of your apparent weights at these two locations? Carry three significant figures in your calculation.
Explanation / Answer
Given: Question Details 1. You are an astronaut (m = 85 kg) and travel to a planet that is the same mass and size as Earth, but it has a rotational period of only 3 h. What is your apparent weight at the equator of this planet?( i have tried the other solutions posted and they don't work, please solve out problem)
2. Your apparent weight is the force you feel on the bottoms of your feet when you are standing. Due to the Earth's rotation, your apparent weight is slightly more when you are at the South Pole than when you are at the equator. What is the ratio of your apparent weights at these two locations SOLUTION: 1. The acceleration of gravity is the same : g = (GM)/(r^2) = 9.8meters/(s^2) . Think of the mass of the astronaut suspended from a rope . The three forces on the body are the uppward tension , the downward gravitational force , and the downward centripital force . We thus have : mg - T = mv^2/r . The tension is the weight that that the astronaut feels on the planet . Sove for the tension , and we get : T = mg - mv^2/r = m( g - v^2/r) (1) The radius of the planet is the same as earth's : r = 6.38(10)^6 meters . We have the speed of rotation at the equator is : v = 2(pi)6.38(10)^6 meters /(3hrs(3600s/hr)) = 3,711.73 meters/s We then have the weight that the astronaut feels at the equator of the planet which is the tension in the rope is using equation (1) : T = 85(9.8 - (3,711.73)^2/(6.38(10)^6)) = 649.45 Newtons ----- solution ------------------------------------------------------------------------------------ His weight on earth would be --- w = 9.8(85) = 833 Newton's . We thus have his weight on the equator of the planet is 77.97 % of his weight on earth. 2. All we have to do is use equation (1) again . Your weight at the south pole is : Ws = mg (2) Your weight at the equator is : We = m(g - v^2/r) where the velocity at the equator is : v = 2(pi)6.38(10)^6 meters/(24hr(3600s/hr)) = 463.96669 meters/s . Your weight at the equator is thus : We = m(g - (463.96669)^2/(6.38(10)^6)) (3) Divide equation (2) by equation (3) , and we get : Ws/We = g/[g - (463.96669)^2/(6.38(10)^6)] = 9.8/[9.8 - (463.96669)^2/(6.38(10)^6)] = 1.00345 ---- solution . --------------------------------- I think you need more than three significant digits for this ratio . We could also write the ratio of your weight at the equator to your weight at the south pole : We/Ws = .996557 which to three significant figures is .997 ---- solution . --------------------------------------------------------------------------------- 1. You are an astronaut (m = 85 kg) and travel to a planet that is the same mass and size as Earth, but it has a rotational period of only 3 h. What is your apparent weight at the equator of this planet?
( i have tried the other solutions posted and they don't work, please solve out problem)
2. Your apparent weight is the force you feel on the bottoms of your feet when you are standing. Due to the Earth's rotation, your apparent weight is slightly more when you are at the South Pole than when you are at the equator. What is the ratio of your apparent weights at these two locations SOLUTION: 1. The acceleration of gravity is the same : g = (GM)/(r^2) = 9.8meters/(s^2) . Think of the mass of the astronaut suspended from a rope . The three forces on the body are the uppward tension , the downward gravitational force , and the downward centripital force . We thus have : mg - T = mv^2/r . The tension is the weight that that the astronaut feels on the planet . Sove for the tension , and we get : T = mg - mv^2/r = m( g - v^2/r) (1) The radius of the planet is the same as earth's : r = 6.38(10)^6 meters . We have the speed of rotation at the equator is : v = 2(pi)6.38(10)^6 meters /(3hrs(3600s/hr)) = 3,711.73 meters/s We then have the weight that the astronaut feels at the equator of the planet which is the tension in the rope is using equation (1) : T = 85(9.8 - (3,711.73)^2/(6.38(10)^6)) = 649.45 Newtons ----- solution ------------------------------------------------------------------------------------ His weight on earth would be --- w = 9.8(85) = 833 Newton's . We thus have his weight on the equator of the planet is 77.97 % of his weight on earth. 2. All we have to do is use equation (1) again . Your weight at the south pole is : Ws = mg (2) Your weight at the equator is : We = m(g - v^2/r) where the velocity at the equator is : v = 2(pi)6.38(10)^6 meters/(24hr(3600s/hr)) = 463.96669 meters/s . Your weight at the equator is thus : We = m(g - (463.96669)^2/(6.38(10)^6)) (3) Divide equation (2) by equation (3) , and we get : Ws/We = g/[g - (463.96669)^2/(6.38(10)^6)] = 9.8/[9.8 - (463.96669)^2/(6.38(10)^6)] = 1.00345 ---- solution . --------------------------------- I think you need more than three significant digits for this ratio . We could also write the ratio of your weight at the equator to your weight at the south pole : We/Ws = .996557 which to three significant figures is .997 ---- solution . ---------------------------------------------------------------------------------
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