Object falling; acceleration based on x

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



For a spherical celestial object of radius R, the acceleration due to gravity g at a distance x from the center of the object is g = (g0)(R^2)/(x^2), where g0 is the acceleration due to gravity at the object's surface and x > R. For the moon, take g0 = 1.63 m/s2 and R = 3200 km. If a rock is released from rest at a height of 3R above the lunar surface, with what speed does the rock impact the moon? Hint: Its acceleration is a function of position and increases as the object falls. So do not use constant acceleration free-fall equations, but go back to basics.

Homework Equations



.5(v^2-v0^2) = (g0)(R^2)(1/x-1/x0)

The Attempt at a Solution


i said that v0=0 and solved for v, x=3R x0=R, i got .899km/s. apparently that is incorrect.
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,866
What's the difference between height and radial distance?
 
  • #3
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2R, so would i plug 2R in as x and 0 for x0?
 
  • #4
gneill
Mentor
20,925
2,866
2R, so would i plug 2R in as x and 0 for x0?

What's the radial distance to the surface of the Moon?

What's the radial distance to an object at height 3R above the Moon's surface?
 
  • #5
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What's the radial distance to the surface of the Moon?

R

What's the radial distance to an object at height 3R above the Moon's surface?

4R

I see now, i think.
so x0=4R, x=R right?
 
  • #6
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i got 2.797km/s is that the correct answer?
 
  • #7
gneill
Mentor
20,925
2,866
It looks good.
 
  • #9
gneill
Mentor
20,925
2,866
Does the system care if you enter the result as km/sec or m/sec?
 
  • #10
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it wants the answer in km/s
 
  • #11
gneill
Mentor
20,925
2,866
Is it fussy about significant figures?
 
  • #12
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i got it, i think i plugged it into the calculator wrong.
 
  • #13
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thanks for all the help
 

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