How Does Apparent Weight Change for an Astronaut Near the Moon?

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SUMMARY

The apparent weight of a 70-kg astronaut 3800 km from the center of the Moon is calculated using the formula wt = m(g+a) for acceleration towards the Moon and wt = m(g-a) for constant velocity. At a constant velocity, the astronaut's weight is 23.7 Newtons, derived from the Moon's gravitational acceleration of 0.33 m/s². When accelerating towards the Moon at 2.9 m/s², the apparent weight is -179.9 Newtons, indicating that the astronaut feels a force pushing them away from the Moon, which is a result of accelerating faster than the gravitational pull. This negative value signifies that the astronaut experiences an external force counteracting gravity.

PREREQUISITES
  • Understanding of gravitational force and acceleration
  • Familiarity with Newton's laws of motion
  • Basic knowledge of physics equations involving mass and acceleration
  • Ability to interpret negative values in physical contexts
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  • Study the concept of apparent weight in non-inertial reference frames
  • Learn about gravitational acceleration variations on different celestial bodies
  • Explore the implications of acceleration greater than gravitational pull in various scenarios
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Abu
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Homework Statement


What is the apparent weight of a 70-kg astronaut 3800 km from the center of the Earth's Moon in a space vehicle when
a) Moving at a constant velocity
b) Accelerating toward the Moon at 2.9 m/s^2?

State direction in each case

Homework Equations


wt = m(g+a)
wt = m(g-a)
Fgr = GmM/r^2

The Attempt at a Solution


So for a) I first found the acceleration due to gravity at that point, which is
g = GM/r^2
The mass of the moon is 7.36x10^22
g = 6.67x10^-11 (7.36x10^22)/3800000
g = 0.33 m/s

At a constant velocity, the weight is simply mg. So, 70x0.33 = 23.7 Newtons

For b however, I am not entirely sure. This is what I was thinking when I was trying to solve the problem:
Since you are accelerating at 2.9m/s^2 towards the moon, I will use wt = m(g-a) because I was taught that when going against the force of gravity, the wt = m(g+a) is used, and vice versa.
I get the following answer:
wt = 70(0.33 - 2.9)
= -179.9 Newtons

Now I have a couple of problems. I am not sure what direction these values for a) and b) are pointed towards and if a negative apparent weight is even possible (or if that answer is even correct in the first place).

Thank you very much for your patience.
 
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Abu said:
am not sure what direction these values for a) and b) are pointed towards and if a negative apparent weight is even possible (or if that answer is even correct in the first place).
Your work looks correct to me.
 
haruspex said:
Your work looks correct to me.
Thank you very much for your response.

I am glad my work checks out, though I can't seem to wrap my head around why the answer for part b is negative..

Included in that is my confusion regarding which direction these apparent weight forces should be pointing:

My first guess is that for a) there really is no apparent weight because the object is moving at a constant velocity, hence the answer I think would be its gravitational force of attraction directed towards the moon.

I am not sure about b) though..

Thank you for your time.
 
Abu said:
why the answer for part b is negative..
Because weight is, by definition, a force due to gravity, so apparent weight would be the apparent force in the direction of the gravitational field. If the astronaut feels as though a mysterious is pushing her away from the moon then the apparent weight is negative.
 
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haruspex said:
Because weight is, by definition, a force due to gravity, so apparent weight would be the apparent force in the direction of the gravitational field. If the astronaut feels as though a mysterious is pushing her away from the moon then the apparent weight is negative.
Thank you for your time, and sorry for the late response - I got caught up in some other subjects...

So is this what occurs when one were to accelerate at a rate greater than the gravity acting on them? It would feel as though they were getting pushed away? If so... why does this happen? I am assuming that this would also apply to if you were somehow accelerating downwards faster than 9.81 m/s^2 while in an elevator, and thus you would be pushed by some force towards the ceiling of the elevator?

Once again, thank you for your responses.
 
Abu said:
It would feel as though they were getting pushed away?
Yes. Gravity would not be enough to make them accelerate at that rate, so an additional force must be pushing them towards the planet. The astronaut would feel that force as an externally applied force. In the astronaut's frame of reference there is no acceleration. To explain that, despite the felt force, it would feel as though some other pervading force, like gravity, were pushing them up against it.
This is just like the everyday experience of gravity, but inverted. You feel the force of the floor on you, but you are not accelerating, so infer the force of gravity pushing you against the floor.
Abu said:
this would also apply to if you were somehow accelerating downwards faster than 9.81 m/s^2 while in an elevator, and thus you would be pushed by some force towards the ceiling of the elevator?
Yes. Indeed, you would not exceed 9.81 m/s^2 until you found yourself pressed against the ceiling.
 
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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