How Does Acceleration Affect an Astronaut's Apparent Weight Near the Moon?

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The apparent weight of a 75-kg astronaut 4200 km from the Moon's center is 21 Newtons directed toward the Moon when moving at constant velocity, as calculated using gravitational force equations. In the case of acceleration at 2.9 m/s² toward the Moon, the astronaut's apparent weight must account for both gravitational force and the additional acceleration. The normal force will be less than the gravitational force due to the downward acceleration, similar to an elevator scenario. Understanding that "apparent weight" refers to the normal force is crucial for solving these problems. The discussion emphasizes the need to combine gravitational calculations with the effects of acceleration to determine the astronaut's apparent weight accurately.
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What is the apparent weight of a 75-kg astronaut 4200 km from the center of the Earth's Moon in a space vehicle (a) moving at constant velocity, and (b) accelerating toward the Moon at 2.9 m/s^2? State the "direction" in each case.

I understand part a. I got 21 Newtons towards the moon by using Fg = G * [ (m1 * m2) / d^2 ].

For part B I do not understand where / how to use the acceleration of 2.9 m/s^2...would it be the same equation?

mass of the moon: 7.35 x 10^22 kg
mass of the astronaut: 75 kg
distance: 4200 km = 4200000 meters
G = 6.67 x 10^-11
 
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N_L_ said:
What is the apparent weight of a 75-kg astronaut 4200 km from the center of the Earth's Moon in a space vehicle (a) moving at constant velocity, and (b) accelerating toward the Moon at 2.9 m/s^2? State the "direction" in each case.

I understand part a. I got 21 Newtons towards the moon by using Fg = G * [ (m1 * m2) / d^2 ].

For part B I do not understand where / how to use the acceleration of 2.9 m/s^2...would it be the same equation?

mass of the moon: 7.35 x 10^22 kg
mass of the astronaut: 75 kg
distance: 4200 km = 4200000 meters
G = 6.67 x 10^-11

Remember that "apparent weight" is a fancy term for "normal force." In the first case the normal force and the weight are going to be the same magnitude because the astronaut is not accelerating. The accelerating case is very similar to an elevator problem, where the elevator is accelerating downward at a rate less than g.

-Dan
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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