- #1
kindlin
- 12
- 1
So, I asked myself the question, "why does the Earth's force of gravity effect us so dramatically on the surface of the earth, but seems nonexistent while just a couple hundred miles up?"
I answered that question myself, after thinking more about it, because the idea of an orbit is that you constantly fall towards the earth. So just like the simulated weightless environment a plane can create, so is a space station a simulated weightless environment.
However, in trying to answer this question I tried to calculate the Newtonian gravitational force one one would feel at the surface, and here's what I got: (m1=earth, m2=unity)
F = G * m1 * m2 / r^2 = 6.67e-11 * 5.97e24 * 1 / (6378.1 / 2)^2 = 39191036.5 N/kg
So if I weighed 80kg, I would experience 39191037*80 = 3135282921 N
So, what did I do wrong? My feet obviously aren't holding up 314e6 N.
Also, knowing that we feel 9.81*m Newtons sitting on the surface, it would follow that G*m1/r^2 = 9.8:
G = 9.8*r^2/m1 = 9.8* (6378.1 / 2)^2 / 5.97e24 = 1.6705e-17
Thoughts?
I answered that question myself, after thinking more about it, because the idea of an orbit is that you constantly fall towards the earth. So just like the simulated weightless environment a plane can create, so is a space station a simulated weightless environment.
However, in trying to answer this question I tried to calculate the Newtonian gravitational force one one would feel at the surface, and here's what I got: (m1=earth, m2=unity)
F = G * m1 * m2 / r^2 = 6.67e-11 * 5.97e24 * 1 / (6378.1 / 2)^2 = 39191036.5 N/kg
So if I weighed 80kg, I would experience 39191037*80 = 3135282921 N
So, what did I do wrong? My feet obviously aren't holding up 314e6 N.
Also, knowing that we feel 9.81*m Newtons sitting on the surface, it would follow that G*m1/r^2 = 9.8:
G = 9.8*r^2/m1 = 9.8* (6378.1 / 2)^2 / 5.97e24 = 1.6705e-17
Thoughts?