You are a tourist on space odyssey; at what altitude above the surface

AI Thread Summary
To determine the altitude at which a tourist's weight is half of their weight on Earth's surface, the gravitational force equation Fg = GMm/r^2 is used. The discussion clarifies that the relevant variable to adjust is the distance from the center of the Earth, R, which includes Earth's radius plus the altitude above the surface. The gravitational constant G and Earth's mass M remain unchanged. The conversation emphasizes that the altitude must be calculated in terms of Earth's radius, rE, and highlights the importance of distinguishing between actual and apparent weight in different gravitational contexts. Ultimately, the solution involves finding the correct altitude where gravitational force is halved.
jasonbans
Messages
45
Reaction score
0

Homework Statement


You are a tourist on space odyssey; at what altitude above the surface of Earth will your weight be one-half your weight on the surface? express your answer as a mutiple of Earth's radius rE


Homework Equations


Fg= GMm/r^2


The Attempt at a Solution


not sure how to do it
 
Physics news on Phys.org


jasonbans said:

Homework Statement


You are a tourist on space odyssey; at what altitude above the surface of Earth will your weight be one-half your weight on the surface? express your answer as a mutiple of Earth's radius rE


Homework Equations


Fg= GMm/r^2


The Attempt at a Solution


not sure how to do it

Do you mean your actual weight or your apparent weight? Remember the apparent weight of people on the space station is zero as they are in free fall, but their actual weight is around 90% of their weight on Earth - because there g ~ 8.9 rather than the 9.8 on the surface.
 


PeterO said:
Do you mean your actual weight or your apparent weight? Remember the apparent weight of people on the space station is zero as they are in free fall, but their actual weight is around 90% of their weight on Earth - because there g ~ 8.9 rather than the 9.8 on the surface.

actual weight
 


jasonbans said:
actual weight

Look at your formula and consider what change can be made to halve the force.

I don't think you can change G - is is after all the Universal Gravitational Constant.

M won't alter - mass of the Earth

You could halve m if you put the passenger on a diet, but I don't think that is feasible.

Doesn't leave much else - and remember R is the distance from the centre of the Earth, so we are already one Earth radius away when standing on the surface.
 
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...
Thread 'A cylinder connected to a hanged 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...
Back
Top