Speed of a projectile launched from the moon?

  • Thread starter gixx3r
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  • #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 projectile's initial speed?

Homework Equations



h = 370,000 m
g moon = 1.62 m/s^2
G = 6.67x10^-11 N m^2/kg^2
Mass moon = 7.35x10^22 kg
Radius moon = 1.74X10^6 m

vf^2 = vi^2 - 2gh

The Attempt at a Solution



I calculated g on the moon to be 1.62 m/s^2. I then used kinematics (vf^2 = vi^2 - 2gh) and I get:

vf = 0
vi = ?
g = 1.62 m/s^2
vf^2 = Vi^2 - 2gh
solving for Vi = sqroot of Vf^2 + 2gh

I get initial velocity to be 1095.57 m/s. The answer is not correct, and I really don't understand what I'm doing wrong.
 

Answers and Replies

  • #2
mgb_phys
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Two methods
Conservation of energy
1/2mv^2 = mgh
Or motion equations:
v^2 = u^2 + 2gs (where final velocity is zero)

ps. g = 1.62m/s^2 for the moon is correct. Check your calculations
 
  • #3
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Two methods
Conservation of energy
1/2mv^2 = mgh
Or motion equations:
v^2 = u^2 + 2gs (where final velocity is zero)

ps. g = 1.62m/s^2 for the moon is correct. Check your calculations

Thanks for your reply. I tried the equations again and I'm still getting the same answer. It is wrong and I don't understand why :(

Based on the conservation of energy equation you mentioned:
1/2mv^2 = mgh
1/2v^2 = gh
v^2 = 2gh
v = sqroot (2gh) = sqroot [(2)(1.62m/s^2)(370000m)]
v = 1095 m/s
 
  • #4
mgb_phys
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Sorry the altitude is 370km so you can't assume that g is constant.
You need to do gravitational potential energy at the surface and at 370km - the difference is the KE

(Energy = GMm /r)
 
  • #5
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Sorry the altitude is 370km so you can't assume that g is constant.
You need to do gravitational potential energy at the surface and at 370km - the difference is the KE

(Energy = GMm /r)

Ok, here is what I got using your advice:

GPE at the surface:
GMm/r = (6.67x10^-11 N m^2/kg^2)(7.35x10^22 kg) / 1.74x10^6 m = 2817500 J

GPE at 370km:
GMm/r = (6.67x10^-11 N m^2/kg^2)(7.35x10^22kg) / 370000m + 1.74x10^6 m = 2323436 J

2817500 J - 2323436 J = 494064 J = KE

Am I correct so far? Do I plug this number into 1/2mv^2 ?
 
  • #6
mgb_phys
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Almost - you can't have Joules because you don't know the mass of the object (essentially you have assumed a 1kg mass) but that doesn't matter since you can also assume 1kg in the KE
 
  • #7
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I plugged in 494064 into KE = 1/2 mv^2

494064 = 1/2 (7.35x10^22kg) v^2
v = 3.66x10^-9

Does this sound correct?
 
  • #8
mgb_phys
Science Advisor
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No, the 'M' in that equation is the mass of the object you are throwing - not the mass of the moon.
 

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