- #1
zeromodz
- 246
- 0
Okay, I just found this new equation (new to me) for the escape velocity of a body.
escape velocity = √(2GM)/R
So say for instance I plug this into the escape velocity of the earth
I am going to say the radius is 4001 miles so I it would take a mile above the Earth to escape (4001 * 1600) = 6401600 meters
escape velocity = √((2)(6.674*10^-11)(6.0*10^24)/(6401600)
= 11185.08934 m/s
This obviously isn't the speed you need to escape the Earth's gravity a mile off of its surface because we have left Earth already with much lower speeds. This is so fast we would get a little time dilation, can someone tell me what I am doing wrong?
Thanks in advanced.
escape velocity = √(2GM)/R
So say for instance I plug this into the escape velocity of the earth
I am going to say the radius is 4001 miles so I it would take a mile above the Earth to escape (4001 * 1600) = 6401600 meters
escape velocity = √((2)(6.674*10^-11)(6.0*10^24)/(6401600)
= 11185.08934 m/s
This obviously isn't the speed you need to escape the Earth's gravity a mile off of its surface because we have left Earth already with much lower speeds. This is so fast we would get a little time dilation, can someone tell me what I am doing wrong?
Thanks in advanced.