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

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Homework Help Overview

The discussion revolves around a problem involving the velocity of a piece of ejecta thrown upwards, with a focus on energy conservation principles in a gravitational context. The subject area includes concepts from mechanics and gravitational physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of energy conservation equations, questioning the correctness of the quoted equations and the implications of using different reference points for radius. There are attempts to clarify the definitions of variables and the physical meaning behind the equations.

Discussion Status

There is an ongoing exploration of the equations and their implications, with some participants providing corrections and others expressing concerns about the physical feasibility of the calculated velocity. The discussion reflects a mix of interpretations and attempts to reconcile different aspects of the problem.

Contextual Notes

Participants note the importance of measuring distances from the center of mass of the moon and question the treatment of potential energy as the ejecta approaches the surface. There is also mention of the potential for significant speed upon impact, raising questions about the physical realism of the calculated values.

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Homework Statement


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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??
 
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nysnacc said:
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.
 
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??
 
nysnacc said:
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...
 
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, because the radius above surface will be 0, making the expression 1/0
 
nysnacc said:
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, because 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.
 
Big R is the radius of the moon
while r is the distance above the surface?
 
nysnacc said:
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.
 
V = 1447.98 m/s ??
 
  • #10
nysnacc said:
V = 1447.98 m/s ??
Seems too much. Please post all your working.
 
  • #11
-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
nysnacc said:
-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
But does it make sense to crash the surface at such speed?
 
  • #14
nysnacc said:
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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