How Does Acceleration Affect an Astronaut's Apparent Weight?

AI Thread Summary
The apparent weight of a 730-N astronaut in a spaceship accelerating at 2.0g just above Earth's surface is calculated to be 2190N, which rounds to 2200N in the textbook. The discrepancy arises from rounding conventions, as the astronaut's weight calculation is based on the assumption of three significant figures. In a scenario far from any gravitational influences, the only force acting on the astronaut is the normal force exerted by the spaceship, simplifying the calculation to N=ma. The discussion emphasizes the importance of significant figures in scientific calculations. Understanding these concepts can aid in accurately comparing results with textbook answers.
mysticbms
Messages
8
Reaction score
0

Homework Statement


What is the apparent weight of a 730-N astronaut when her spaceship has an acceleration of magnitude 2.0g in the following two situations. a) just above the surface of Earth, acceleration straight up; b) far from any stars of planets?


Homework Equations


Fnet=N-mg=ma


The Attempt at a Solution


W'=N=mg+ma=m(g+a)
=m(g + 2g)
m=730/g
W'=730*3=2190N

The answer in the book is 2200N, not sure if they just rounded and I'm not sure how to answer b.
 
Physics news on Phys.org
mysticbms said:

Homework Statement


What is the apparent weight of a 730-N astronaut when her spaceship has an acceleration of magnitude 2.0g in the following two situations. a) just above the surface of Earth, acceleration straight up; b) far from any stars of planets?

Homework Equations


Fnet=N-mg=ma

The Attempt at a Solution


W'=N=mg+ma=m(g+a)
=m(g + 2g)
m=730/g
W'=730*3=2190N

The answer in the book is 2200N, not sure if they just rounded and I'm not sure how to answer b.
Yup. They rounded ... apparently to two sig fig
 
This book is killing me. Not the first time it had me thinking I got the wrong answer.

What about b?
 
b is actually simpler than a. The reason the book tells you that the spaceship is far from any planets or stars is to tell you that there is a negligible amount of gravitational force acting on the ship. Therefore, the only force acting on the astronaut in this situation is the normal force that the spaceship exerts on the astronaut. So instead of having to add the normal force and gravity, it's just the normal force that makes up the net force.

So in that case, N=ma
 
mysticbms said:
This book is killing me. Not the first time it had me thinking I got the wrong answer.
In a sense you do have the wrong answer. You only know the acceleration to two places. It's 2.0 g, not 2.00g. Giving too much precision in an answer is a wrong answer.
 
D H said:
In a sense you do have the wrong answer. You only know the acceleration to two places. It's 2.0 g, not 2.00g. Giving too much precision in an answer is a wrong answer.

Thank you! I didn't think of it that way. That will definitely help me moving forward when comparing answers to the book.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

How Does Apparent Weight Change for an Astronaut Near the Moon?
Replies
5
Views
3K
Speed and apparent weight of person in reverse bungee jumping
Replies
9
Views
2K
True vs. Apparent weight question
Replies
10
Views
4K
Astronaut's Apparent Weight in Different Gravitational Fields
Replies
1
Views
2K
Calculating the Ratio of Apparent Weight on a Ferris Wheel
Replies
11
Views
7K
Apparent weight problem (kinematics + conservation of Energy + Newton's laws)
Replies
41
Views
3K
What is the apparent weight of a woman riding over a hump in a car?
Replies
6
Views
3K
Apparent weight of a man standing in an elevator
Replies
24
Views
7K
Calculating Apparent Weight on a Rotor Ride
Replies
3
Views
2K
Calculating Mass and Weight on the Moon
Replies
4
Views
3K

Hot Threads

Recent Insights

Back
Top