A piece of ejecta is thrown up, what is the velocity

  • Thread starter nysnacc
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  • #1
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Homework Statement


upload_2016-10-6_14-8-29.png

upload_2016-10-6_14-8-43.png

upload_2016-10-6_14-9-40.png


Homework Equations


energy conservation

The Attempt at a Solution


I used this equation:

1/2 mg R^2/ r1 + 1/2 mv1^2 = 1/2 mg R^2/ r2 + 1/2 mv2^2
For r1, it is 1738 km (radius) + 1000 km (above surface) ?
But for r2 , it is on the surface, so would it be 0??
 

Answers and Replies

  • #2
haruspex
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1/2 mg R^2/ r1
That does not look quite right to me. Are you sure you have quoted it correctly?
Also you need to think about the sign.
 
  • #3
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Sorry.
mg R^2/ r1 + 1/2 mv1^2 = mg R^2/ r2 + 1/2 mv2^2
For r1, it is 1738 km (radius) + 1000 km (above surface) ?
But for r2 , it is on the surface, so would it be 0??
 
  • #4
haruspex
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Sorry.
mg R^2/ r1 + 1/2 mv1^2 = mg R^2/ r2 + 1/2 mv2^2
For r1, it is 1738 km (radius) + 1000 km (above surface) ?
But for r2 , it is on the surface, so would it be 0??
I edited my first reply while you were responding...
 
  • #5
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So it would be - before mg? based on the definition?

But my concern is mainly what happened with the potiential energe when it hits the surface, cuz the radius above surface will be 0, making the expression 1/0
 
  • #6
haruspex
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So it would be - before mg? based on the definition?

But my concern is mainly what happened with the potiential energe when it hits the surface, cuz the radius above surface will be 0, making the expression 1/0
The equations you quote all take radii as being measured from the centre of mass of the moon.
 
  • #7
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Big R is the radius of the moon
while r is the distance above the surface?
 
  • #8
haruspex
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Big R is the radius of the moon
while r is the distance above the surface?
You are given the distance above the surface, but the equations you quote require r to be measured from the mass centre of the gravitational body.
 
  • #9
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V = 1447.98 m/s ??
 
  • #11
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-mg R2/ r1 +1/2 m v12 = -mg R2/r2 +1/2 m v22

r1 is radius + distance above > R , r2 = R

so
-g R2/ r1 +1/2 v12 = -g R +1/2 v22

With v1 = 200 m/s, g = 1.62 m/s2

-2g R2/ r1 + 2g R +v12 = v22

-2g {R2/ r1 - R} + v12 = v22

-2(1.62) (17380002/2738000 - 1738000) + (200)2 = v22

So then v2 is sqrt of the left hand side, = 1447.98 m/s
 
  • #12
haruspex
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-mg R2/ r1 +1/2 m v12 = -mg R2/r2 +1/2 m v22

r1 is radius + distance above > R , r2 = R

so
-g R2/ r1 +1/2 v12 = -g R +1/2 v22

With v1 = 200 m/s, g = 1.62 m/s2

-2g R2/ r1 + 2g R +v12 = v22

-2g {R2/ r1 - R} + v12 = v22

-2(1.62) (17380002/2738000 - 1738000) + (200)2 = v22

So then v2 is sqrt of the left hand side, = 1447.98 m/s
Fair enough... my rough estimate must have had a mistake somewhere. The 200m/s turns out to be insignificant.
 
  • #13
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But does it make sense to crash the surface at such speed?
 
  • #14
haruspex
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But does it make sense to crash the surface at such speed?
Quick lower bound: up to 1738km from surface, g is at least 1.6/22=0.4m/s2. Using v2=2as=800,000m2/s2, v is at least 900m/s.
 

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