# Homework Help: Projectile Motion in Bigger Radius.

1. Nov 13, 2008

### leyyee

1. The problem statement, all variables and given/known data
A projectile of mass, m is fired from the surface of the earth at an angle alpha from the vertical. THe initial speed v0 is equal to (GM/R)^0.5 where G is gravitational cont, M is mass of earth and R is radius of the earth. How high does the projectile rise? Neglect air resistance and the earth's rotation. r.max=r max in picture.

http://img370.imageshack.us/my.php?image=quesgf6.jpg

http://img370.imageshack.us/img370/189/quesgf6.jpg [Broken]
http://g.imageshack.us/img370/quesgf6.jpg/1/ [Broken]

2. Relevant equations

U + K = constant

3. The attempt at a solution

So I started with that equation, and find out that
-(GMm/R) + .5(m)(v0^2) = -(GMm/(r.max)) + .5(m)(vf^2)

and from here we can eliminate the vf because it is zero when it is a the highest position. But after substitution that v0 = (GM/R)^0.5 this equation will become

-(GMm/R) + .5(GMm/R) = -(GMm/(r.max))

this will become r.max =2R which i think is impossible.

Can anyone out there here me ? I am totally clueless. From the lecturer I found out that is it max when alpha is = 60degrees and the r.max is (3R/2) .

Last edited by a moderator: May 3, 2017
2. Nov 13, 2008

### unscientific

loss in Ke = Gain in GPE

h = distance from the centre of the earth

(1/2)m(v0 cos a)^2 = GMm/R - GMm/h

(1/2)(m)(GM/R) (cos a)^2 = GMm ( 1/R - 1/h)

Cancelling m and GM on both sides,

(1/2R) (cos a)^2 = 1/R - 1/h

1/h = 1/R - (1/2R) (cos a)^2

1/h = (2 - cos^2 a)/2R

h = (2R)/(sin^2 a + 1)

Therefore, for a maximum h, sin a must be as small as possible, setting alpha = 0 degrees gives h = 2R.

when alpha = 60 degrees, h = (2R)/(1.75) = 1.14R

Last edited: Nov 13, 2008
3. Nov 13, 2008

### leyyee

hey unscientific as I know that from the lecturer it is a round number.. Could it be I have done wrong somewhere? I got the same answer as well but I just want to confirm again?

Anyone else might wanna try to help me?

Thanks..