The discussion centers on determining the altitude above Earth's surface where a tourist's weight is half of their weight on the surface. The relevant formula is Fg = GMm/r², where Fg represents gravitational force, G is the universal gravitational constant, M is Earth's mass, m is the mass of the tourist, and r is the distance from the center of the Earth. To achieve half the weight, the distance r must be adjusted, taking into account that standing on the surface already places the tourist one Earth radius (rE) away from the center. The conversation emphasizes that altering G or M is not feasible, and practical solutions are limited to changing r.
PREREQUISITESStudents studying physics, educators teaching gravitational concepts, and anyone interested in the effects of altitude on weight in space travel scenarios.
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
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.
jasonbans said:actual weight