explain the concept of escape velocity (1 Viewer)

[sadpwner]

explain the concept of escape velocity
in terms of the:
– gravitational constant
– mass and radius of the planet

I know the definition of escape velocity, but how would you explain it in those terms? I read that the equation for escape velocity v=sqr(2GM/r) is not part of the syllabus.

 

[InteGrand]

Escape velocity is the velocity given by , where M is the mass of the planet, and R_planet is the radius of the planet. Thus as you can see, the square root of the gravitational constant helps determine v_esc., and v_esc. is proportional to the square root of the mass of the planet, and inversely proportional to the square root of the planet's radius.

Here is the explanation why (i.e. how we derived the formula for v_esc.).

We are assuming there's no loss of energy from the spacecraft due to air resistance or other frictional effects etc. Therefore, the total mechanical energy (which is E_tot. = E_k + E_p) of the craft is conserved throughout its flight (conservation of energy). This means the value of E_tot. = E_k + E_p is constant at any point in time of the craft's flight. As we are launching it at escape velocity, we want in the limit at time goes to infinity, the distance the craft is from the planet to tend to infinity (this is the definition of the escape velocity: the craft will get arbitrarily far away from the planet as time goes on and on – it doesn't reach a maximum distance and then fall back down to Earth), and in this limit, the speed of the spacecraft will tend to 0. This means in the limit as , we have (because the velocity is tending to 0 as the earth's gravity keeps slowing the craft down) and (because the distance r of the craft tends to infinity, and as r is in the denominator of the E_p formula, ). This means the sum of kinetic energy and potential energy (i.e. total mechanical energy), in the limit as , is 0.

Since , and we said earlier that by conservation of energy, is constant, it means this constant value is 0, i.e. .

, since when time = 0, the craft starts at the planet's surface (hence E_p has R_planet in the denominator, as this is the initial distance from the planet's centre), and the velocity is the escape velocity (since we're launching it at escape velocity), so E_k initially equals (m is the mass of the craft).

So that's the explanation of how we got the equation at the start, from which we derived escape velocity's formula.

 

[mreditor16]

Here's my explanation of it (from my notes) -

 

[InteGrand]

Here's my explanation of it (from my notes) -

KE doesn't equal GPE, it equals negative GPE (because (reason for this is explained in my above post)).

 

[QZP]

Escape velocity is the velocity given by , where M is the mass of the planet, and R_planet is the radius of the planet. Thus as you can see, the square root of the gravitational constant helps determine v_esc., and v_esc. is proportional to the square root of the mass of the planet, and inversely proportional to the square root of the planet's radius.

Here is the explanation why (i.e. how we derived the formula for v_esc.).

We are assuming there's no loss of energy from the spacecraft due to air resistance or other frictional effects etc. Therefore, the total mechanical energy (which is E_tot. = E_k + E_p) of the craft is conserved throughout its flight (conservation of energy). This means the value of E_tot. = E_k + E_p is constant at any point in time of the craft's flight. As we are launching it at escape velocity, we want in the limit at time goes to infinity, the distance the craft is from the planet to tend to infinity (this is the definition of the escape velocity: the craft will get arbitrarily far away from the planet as time goes on and on – it doesn't reach a maximum distance and then fall back down to Earth), and in this limit, the speed of the spacecraft will tend to 0. This means in the limit as , we have (because the velocity is tending to 0 as the earth's gravity keeps slowing the craft down) and (because the distance r of the craft tends to infinity, and as r is in the denominator of the E_p formula, ). This means the sum of kinetic energy and potential energy (i.e. total mechanical energy), in the limit as , is 0.

Since , and we said earlier that by conservation of energy, is constant, it means this constant value is 0, i.e. .

, since when time = 0, the craft starts at the planet's surface (hence E_p has R_planet in the denominator, as this is the initial distance from the planet's centre), and the velocity is the escape velocity (since we're launching it at escape velocity), so E_k initially equals (m is the mass of the craft).

So that's the explanation of how we got the equation at the start, from which we derived escape velocity's formula.

Quite lengthy... I prefer to explain it as:

Definition: Escape velocity is the minimum initial speed required for an object to have zero (negligible) gravitational attraction to a massive body without further propulsion (i.e. the object would have 0 KE and 0 GPE after 'escaping' the gravitational field).

Hence, KE + GPE = 0 from the conservation of mechanical energy and above definition (KE + GPE is constant and KE = 0 + GPE = 0 from the above scenario, therefore always KE + GPE = 0).

*Derive escape velocity in like 2-3 lines as normal*

 

[InteGrand]

Quite lengthy... I prefer to explain it as:

Definition: Escape velocity is the minimum initial speed required for an object to have zero (negligible) gravitational attraction to a massive body without further propulsion (i.e. the object would have 0 KE and 0 GPE after 'escaping' the gravitational field).

Hence, KE + GPE = 0 from the above definition and the conservation of mechanical energy.

*Derive escape velocity in like 2-3 lines as normal*

It can't escape the gravitational field though. "Escaping" is taken to mean distance tending to infinity, which is what I said.

And what do you mean further propulsion? Its gravitational attraction to Earth is only dependent on distance, not speed or launch speed.

 

[QZP]

It can't escape the gravitational field though. "Escaping" is taken to mean distance tending to infinity, which is what I said.

And what do you mean further propulsion? Its gravitational attraction to Earth is only dependent on distance, not speed or launch speed.

Yes I put quotation marks around 'escaping' because I was using the definition I provided in the previous line. It means to reach a point of zero (negligible) gravitational attraction to a massive body.

What do I mean by without further propulsion? I can launch a spacecraft directly upwards with a constant velocity of 5 m/s. Will it eventually "escape"? Obviously. Is 5 m/s its escape velocity? Obviously not. Escape velocity is the speed you need to initially launch it at without further propulsion to just reach the point of zero (negligible) gravitational attraction from the massive body.

 

[anomalousdecay]

KE doesn't equal GPE, it equals negative GPE (because (reason for this is explained in my above post)).

Although correct for the situation above as mechanical energy is conserved, be very very careful with how you use this equation.

If non-conservative forces do work (friction, drag, etc), then mechanical energy is NOT conserved and hence

This is something learnt mainly in uni, but it is a good thing to note down that if friction or drag is involved and hence mechanical energy is not conserved, then you can't equate that to zero.

 

[mreditor16]

KE doesn't equal GPE, it equals negative GPE (because (reason for this is explained in my above post)).

Yep, my bad. Should have clarified that by that I meant magnitude only!