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A projectile is shot straight up from the Earth's surface 
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#1
Apr1613, 12:19 PM

P: 37

1. The problem statement, all variables and given/known data
A projectile is shot straight up from the earth's surface at a speed of 1.10×10^4 km/hr 2. Relevant equations mgh=.5mv^2 3. The attempt at a solution I converted the speed to m/s and 3055.5556 m/s masses cancel out so i get 9.8h=.5v^2 I plugged in v and solved for H and got 476347.95m, which was wrong. I'm not sure if this is how you are supposed to solve this problem, but i can't think of any other ways. 


#2
Apr1613, 12:28 PM

Thanks
P: 5,801

"mgh" works only for low altitudes.



#3
Apr1613, 12:39 PM

P: 37

What other gravitational equations have height in them? 


#4
Apr1613, 12:52 PM

Thanks
P: 5,801

A projectile is shot straight up from the Earth's surface
Newton's law of gravitation?



#5
Apr1613, 01:01 PM

P: 37

Newton's law of gravitation is Fg=GmM/r^2. or Ug= GmM/r
how would I get height out of that? 


#6
Apr1613, 01:03 PM

Thanks
P: 5,801

What is "r"?



#7
Apr1613, 01:11 PM

P: 37

radius and I'm not given any masses. the m would cancel out if I do KE=PE, but the M wouldnt



#8
Apr1613, 01:29 PM

Thanks
P: 5,801

Radius from where?



#9
Apr1613, 01:35 PM

P: 37

r would be radius from center of the earth to the projectile.
so would I do GmM/R=.5mv^2? G=6.673x10^11 M=5.972x10^24 solve for R and subtract by radius of earth? 


#10
Apr1613, 01:42 PM

Thanks
P: 5,801

That would be correct. Note that you could find out what GM is by considering the magnitude of the force of gravity right at the surface of the Earth.



#11
Apr1613, 01:52 PM

P: 37

I converted 1.10x10^4 km/hr to 3055.55555m/s and solved for R and got 85328725.7m and then I subtracted by the Earth Radius which is 6371000m and got 78955725.7 as the final answer but that is still wrong



#12
Apr1613, 02:30 PM

Mentor
P: 11,689

The projectile begins at a certain radius, so it has some initial gravitational potential. It ends up at some new distance with another gravitational potential. The KE from the launch is exchanged for this change in gravitational PE.
So you should write an expression for the change in gravitational PE and relate it to the change in KE. 


#13
Apr1613, 06:04 PM

P: 552

Excuse me for butting in but isn't there a relationship something like E = G(m1)(m2)(1/(r1) 1/(r2))
or you can use calculus to integrate E = Int(G(m1)(m2)/r^2) from the surface of the earth r1 to r2 


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