- #1
jcais
- 22
- 0
Hello all,
I was wondering if someone could please help me with a problem. It is regarding the law of universal gravitation, this time using conservation of mechanical energy.
We know that F = GMm/R^2
**If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is 27% of the escape speed.**
I approached this problem two ways and have stopped in the middle, because I don't know how to continue. I was wondering if someone can help me find h or the height in meters.
First thing I tried:
.5mv^2 - GmM/Radius of Earth = .5mv^2 - GmM/h
0 = GmM/Radius of Earth - GmM/h goes to h = Radius of earth
I stopped here
Second thing I tried:
.5mv^2- GmM/Radius of Earth = .5m(.27v)^2- GmM/Radius of Earth + h
I do not know how to manipulate this problem to get radius of Earth + h
Thank you for your assistance!
I was wondering if someone could please help me with a problem. It is regarding the law of universal gravitation, this time using conservation of mechanical energy.
We know that F = GMm/R^2
**If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is 27% of the escape speed.**
I approached this problem two ways and have stopped in the middle, because I don't know how to continue. I was wondering if someone can help me find h or the height in meters.
First thing I tried:
.5mv^2 - GmM/Radius of Earth = .5mv^2 - GmM/h
0 = GmM/Radius of Earth - GmM/h goes to h = Radius of earth
I stopped here
Second thing I tried:
.5mv^2- GmM/Radius of Earth = .5m(.27v)^2- GmM/Radius of Earth + h
I do not know how to manipulate this problem to get radius of Earth + h
Thank you for your assistance!