Projectile on the moon

  • Thread starter ryty
  • Start date
  • #1
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Homework Statement


A projectile launched vertically from the surface of the Moon rises to an altitude of 370 km. What was the projectiles initial speed?


Homework Equations


Vf ^2 = Vi^2 + 2ad


The Attempt at a Solution


distance = 370km x 1000 = 3.7E 5 meters
solving for Vi = sqroot of Vf^2 - 2ad
sqrt of 0 - -1.63 x 3.7E5 = 777m/s
 

Answers and Replies

  • #2
ideasrule
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Check your calculations. 1.63 m/s^2 is equal to a, not 2a.
 
  • #3
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alright, i tried this vi=sqrt(2(1.63)(3.7)=1098, but thats wrong too, i still cant find out what im doing wrong
 
  • #4
ideasrule
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1098 m/s is correct, unless the problem wanted you to take into consideration differences in acceleration due to the object's increasing altitude.
 
  • #5
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i dont know, my hw is online, so i just type in an answer, and it ells me whether im right or wrong, and it said i was wrong. So how would i compensate for the difference?
 
  • #6
ideasrule
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You use the conservation of energy: initial kinetic energy + initial potential energy = final kinetic energy + final potential energy. Make sure to use U=-GMm/r for potential energy, not U=mgh.
 
  • #7
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what masses should i use?
 
  • #8
ideasrule
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Well, in U=-GMm/r, M and m are the two attracting masses. Here, one of them is the Moon and the other is the projectile. You'll find that the mass of the projectile cancels out, so you don't need to know what it is.
 
  • #9
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for radius, do i add in radius of the moon, or just use 370km as the radius?
 
  • #10
ideasrule
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"r" is the distance from the center of the planet/moon to the projectile. So yes, you add the radius of the moon.
 
  • #11
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i got 1.01 m/s, which is wrong
0 + G(6.67E-11*7.36E22)/21074000^2 = 1/2vi^2 + G(6.67E-11*7.36E22)/1737400^2
 
  • #12
2,034
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It's U=-GMm/r not U=-GMm/r^2 f
 
  • #13
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just for easier ways to put it Kf=a Uf=b Ki=c Ui=d
it should look like this
c= -b + d
 

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