When two planets collide what is their terminal velocity?

[Janus]

It is just the difference in the escape velocities from the two locations Terminal Impact velocity (Vt)

Vt = sqrt(2*M*G/r1) - sqrt(2*M*G/r2)

M current mass of Earth

r1 - radius of the Earth (current)

R2 distance out to center point of the L4 lagrangian Point which by geometry = radius of the Earth's orbit.

having just glanced through this thread, I can upon this and need to point out an error.

 

Your equation for Vt is wrong. You can't just take the difference of the escape velocities.

 

The way to approach this is by conservation of energy.

 

The total energy of our object with respect to the Earth is equal to the sum of its KE and GPE or

 

[math]\frac{v^2m}{2}-\frac{GMm}{r}[/math]

M is the mass of the Earth and m the mass of our object.

 

At r2, we assume that it starts with a v of 0, so this reduces to

 

[math]-\frac{GMm}{r2}[/math]

 

At the Earth's surface(r1) this will be

 

[math]\frac{Vt^2m}{2}-\frac{GMm}{r1}[/math]

 

Since the total energy does not change, we can say

 

[math]\frac{Vt^2m}{2}-\frac{GMm}{r1}= -\frac{GMm}{r2}[/math]

 

The "m"s cancel out and solving for Vt gives us

 

[math]Vt = \sqrt{2GM \left ( \frac{1}{r1}-frac{1}{r2} \right )}[/math]

 

Note that is a square root of a difference rather than the difference of two square roots.

 

So for instance, with a 150 million km value for r2, using the difference between escape velocities gives an answer of

 

~11131.86 m/s or ~ 73 m/s less than escape velocity from the Earth.

 

The correct method however gives

 

~11204.67 or only ~ 0.24 m/s less than Earth escape velocity.

 

IOW, falling from 150,000,000 km vs infinity results in a fairly insignificant difference in impact speed.

[Robittybob1]

having just glanced through this thread, I can upon this and need to point out an error.

 

Your equation for Vt is wrong. You can't just take the difference of the escape velocities.

 

The way to approach this is by conservation of energy.

 

The total energy of our object with respect to the Earth is equal to the sum of its KE and GPE or

 

[math]\frac{v^2m}{2}-\frac{GMm}{r}[/math]

M is the mass of the Earth and m the mass of our object.

 

At r2, we assume that it starts with a v of 0, so this reduces to

 

[math]-\frac{GMm}{r2}[/math]

 

At the Earth's surface(r1) this will be

 

[math]\frac{Vt^2m}{2}-\frac{GMm}{r1}[/math]

 

Since the total energy does not change, we can say

 

[math]\frac{Vt^2m}{2}-\frac{GMm}{r1}= -\frac{GMm}{r2}[/math]

 

The "m"s cancel out and solving for Vt gives us

 

[math]Vt = \sqrt{2GM \left ( \frac{1}{r1}-frac{1}{r2} \right )}[/math]

 

Note that is a square root of a difference rather than the difference of two square roots.

 

So for instance, with a 150 million km value for r2, using the difference between escape velocities gives an answer of

 

~11131.86 m/s or ~ 73 m/s less than escape velocity from the Earth.

 

The correct method however gives

 

~11204.67 or only ~ 0.24 m/s less than Earth escape velocity.

 

IOW, falling from 150,000,000 km vs infinity results in a fairly insignificant difference in impact speed.

Where did the [math]\frac{Vt^2m}{2}[/math] bit come from for the Earth?

 

Could you please go over this part with more detail to the algebra so I can see how you get there please?

"At r2, we assume that it starts with a v of 0, so this reduces to

[math]-\frac{GMm}{r2}[/math]

 

At the Earth's surface(r1) this will be

[math]\frac{Vt^2m}{2}-\frac{GMm}{r1}[/math]

 

The "m"s cancel out and solving for Vt gives us

[math]Vt = \sqrt{2GM \left ( \frac{1}{r1}-frac{1}{r2} \right )}[/math]"

 

Some of the problem might just be the way the formulas displayed Vt = Terminal velocity OK

and the "frac" bit is a mistake in the way the fraction 1/r2 should display.

Edited April 7, 2015 by Robittybob1

[Janus]

Where did the [math]\frac{Vt^2m}{2}[/math] bit come from for the Earth?

If Vt is to be considered the impact velocity, then using the kinetic energy formula

E = mv^2/2 where m is the mass of the falling object, substituting Vt for v gives the KE of the falling object at impact.

 

 

 

 

 

Could you please go over this part with more detail to the algebra so I can see how you get there please?

"At r2, we assume that it starts with a v of 0, so this reduces to

[math]-\frac{GMm}{r2}[/math]

 

 

 

 

It is obvious that if v=o at r2, then the kinetic energy part of of the total energy equation becomes zero and only leaves the gravitational potential energy at r2 to consider.

 

 

 

 

 

At the Earth's surface(r1) this will be

[math]\frac{Vt^2m}{2}-\frac{GMm}{r1}[/math]

 

 

 

 

Again, this is the sum the KE and GPE of the object at the moment of contact.

 

 

 

 

 

The "m"s cancel out and solving for Vt gives us

[math]Vt = \sqrt{2GM \left ( \frac{1}{r1}-\frac{1}{r2} \right )}[/math]"

 

 

 

 

Starting with the requirement that the total energy of our object with respect to the Earth must remain the same throughout its fall (it is simply exchanging GPE for KE so the sum remains the same), it follows that we can equate the total energy at r2 to the total energy at impact.

 

[math]\frac{Vt^2m}{2}-\frac{GMm}{r1}= -\frac{GMm}{r2}[/math]

 

If you factor out the "m" in the terms on the left, and divide both sides by m, you get m/m =1 on both sides and you have eliminated the 'm's from the equation leaving you with

 

[math]\frac{Vt^2}{2}-\frac{GM}{r1}= -\frac{GM}{r2}[/math]

 

Subtract -GM/r1 from both sides

 

[math]\frac{Vt^2}{2}= -\frac{GM}{r2}+\frac{GM}{r1}[/math]

 

Reverse the order of terms on the right.

 

[math]\frac{Vt^2}{2}= \frac{GM}{r1}-\frac{GM}{r2}[/math]

 

Multiply both sides by 2

 

[math]Vt^2= 2 \left (\frac{GM}{r1}-\frac{GM}{r2} \right )[/math]

 

Factor out GM from both terms on the right

 

[math]Vt^2= 2GM \left (\frac{1}{r1}-\frac{1}{r2} \right )[/math]

 

Take the square root of both sides

 

[math]Vt= \sqrt{ 2GM \left (\frac{1}{r1}-\frac{1}{r2} \right )}[/math]

 

Giving the impact velocity for an object dropped to the surface of the Earth from a distance of r2.

Edited April 8, 2015 by Janus

[imatfaal]

 

I couldn't be bothered with Latex

[Robittybob1]

Very good and thanks to you both. I can follow that perfectly now.

 

1. Check your numbers. You have done an arithmetic error in Excel. For example at L3

 

[latex] v_e = \sqrt \frac {2GM}{r} = \sqrt \frac {2*6.67*10^{-11}*6*10^{24}}{3*10^{11}} = \sqrt \frac {80.04*10^{13}}{3*10^{11}} = \sqrt {26.68*10^2}= 51.65 \, m/s[/latex]

 

The rest of your numbers have the same error.

 

2. Explain physical reason for subtracting square roots of escape velocities at different distances? Can you show me where you got that formula from or explain what it's used for?

 

3. Have you invented the term "Terminal Impact velocity"?

Thanks I had converted the 150 million km incorrectly to meters. My mistake.

 

My answers should have been 11185.8, 72.899, 11112.9

9687.2, 63.132 , 9624.0

But we will need to look at the work by Janus and make other corrections.

Edited April 7, 2015 by Robittybob1

[Robittybob1]

....

[math]Vt= sqrt{ 2GM \left (\frac{1}{r1}-\frac{1}{r2} \right )}[/math]

 

Giving the impact velocity for an object dropped to the surface of the Earth from a distance of r2.

The equation produces reliable results. Thanks.

Edited April 8, 2015 by Robittybob1