escape velocity (1 Viewer)

[felixcthecat]

hi.. i just came across a question while reviewing the topic Space..

this is about escape velocity...

1. when a planet with the same density as earth, but with a bigger volume, escape velocity increase
2. when a planet with the same mass as Earth, but larger radius, escape velocity decrease
but..
1. when a planet with more density than earth, but smaller volume, would its escape velocity increase or decrease?
2. when a planet has a greater mass than Earth, and a bigger radius, would its escape velocity increase or decrease?

anyone kno plz tell me thx!!

 

[eX-Bhai]

hi.. i just came across a question while reviewing the topic Space..

this is about escape velocity...

1. when a planet with the same density as earth, but with a bigger volume, escape velocity increase
2. when a planet with the same mass as Earth, but larger radius, escape velocity decrease
but..
1. when a planet with more density than earth, but smaller volume, would its escape velocity increase or decrease?
2. when a planet has a greater mass than Earth, and a bigger radius, would its escape velocity increase or decrease?

anyone kno plz tell me thx!!

im not too sure on 1. but for 2, it really depends, if the mass of the planet is of a greater proportion than the radius, than the escape velocity would increase. If it was the same proportion, it should increase, but don't take my word on that.

 

[zeropoint]

hi.. i just came across a question while reviewing the topic Space..

this is about escape velocity...

1. when a planet with the same density as earth, but with a bigger volume, escape velocity increase
2. when a planet with the same mass as Earth, but larger radius, escape velocity decrease
but..
1. when a planet with more density than earth, but smaller volume, would its escape velocity increase or decrease?
2. when a planet has a greater mass than Earth, and a bigger radius, would its escape velocity increase or decrease?

anyone kno plz tell me thx!!

As eX-Bhai correctly pointed out, the question is kinda ambiguous, because they don't tell you the proportions by which the variables increase or decrease. It's still possible to arrive at some general results using dimensional analysis, however:

You can tell from the equation

g = GM / r^2

that g is directly proportional to M and inversely proportional to r^2. From the definition of density M = rho * V where rho is the density. So g has the following dependence

g ~ ( rho * V ) / r^2

Note also that V ~ r^3 (remember the formula for the volume of a sphere), so r ~ V^(1/3), and finally

g ~ (rho * V) / (V^2/3) ~ rho * V^(1/3)

so g has linear dependence on density and cube root dependence on volume. This means that if we increase density and decrease volume, gravity increases (linear beats cube root). Stronger gravity then means higher escape velocity

2.

This one follows straight from the formula, g= GM / r^2 increasing mass and increasing radius means decreasing gravity (quadratic beats linear).

 

[felixcthecat]

As eX-Bhai correctly pointed out, the question is kinda ambiguous, because they don't tell you the proportions by which the variables increase or decrease. It's still possible to arrive at some general results using dimensional analysis, however:

You can tell from the equation

g = GM / r^2

that g is directly proportional to M and inversely proportional to r^2. From the definition of density M = rho * V where rho is the density. So g has the following dependence

g ~ ( rho * V ) / r^2

Note also that V ~ r^3 (remember the formula for the volume of a sphere), so r ~ V^(1/3), and finally

g ~ (rho * V) / (V^2/3) ~ rho * V^(1/3)

so g has linear dependence on density and cube root dependence on volume. This means that if we increase density and decrease volume, gravity increases (linear beats cube root). Stronger gravity then means higher escape velocity

2.

This one follows straight from the formula, g= GM / r^2 increasing mass and increasing radius means decreasing gravity (quadratic beats linear).

ummm zeropoint, wut does that mean in english? O_0'' not literally but i havn't met the physics friend, 'rho' and wuts '~' represent?

and for number 2.. the 'quadratic beats linear' i guess makes sense, but only depends on the magnitude it is greater in. Say if M=mass of earth and r=radius of earth, then if the planet had 4M and and 2r, then there would be no change, or if you increase to 5M and 2r, then the g increases.. umm, does that translate to an increase or decrease in escape velocity? (i'm just getting a bit confused with 'g' and esc. velocity)

and true, i guess it's kinda ambiguous ^^''

 

[zeropoint]

ummm zeropoint, wut does that mean in english? O_0'' not literally but i havn't met the physics friend, 'rho' and wuts '~' represent?

and for number 2.. the 'quadratic beats linear' i guess makes sense, but only depends on the magnitude it is greater in. Say if M=mass of earth and r=radius of earth, then if the planet had 4M and and 2r, then there would be no change, or if you increase to 5M and 2r, then the g increases.. umm, does that translate to an increase or decrease in escape velocity? (i'm just getting a bit confused with 'g' and esc. velocity)

and true, i guess it's kinda ambiguous ^^''

I've used ~ to indicate direct proportionality, because I can't find the usual symbol. So, the weight of an object is directly proportional to its mass would be written W ~ m. The greek letter ρ (rho) is used to denote density.

As for number 2, you're right. Like I said the question doesn't specify the proportions of increase/decrease, so it's only possible to speak generally for this question. If you want to find out what happens for specific cases, I would suggest the following routine:

First figure out the proportionality, which in this case is obviously g ~ M / r^2.

Now replace the variables by their increased values (in this case 5M and 2r), and find the new value of g:

g' ~ (5M) / (2r)^2 ~ ( 5/4 ) ( M / r^2 )

which means g' = (5/4)g, i.e. gravity increases.

 

[felixcthecat]

ohh, so if it increases to 5M and 2r, then since the gravitation decreases, the escape velocity would essentially decrease!! oohh i finally linked 'g' with 'escape velocity'!! yay, thanks. lolz

edit: ohh and i think i remember my friend rho now!! i'll hav to kno rho a bit better tho ^^''

 

[zeropoint]

ohh, so if it increases to 5M and 2r, then since the gravitation decreases, the escape velocity would essentially decrease!! oohh i finally linked 'g' with 'escape velocity'!! yay, thanks. lolz

edit: ohh and i think i remember my friend rho now!! i'll hav to kno rho a bit better tho ^^''

The other way around, if you increase to 5M and 2r, gravity increases by a factor of 5/4, which means escape velocity also increases.

 

[zeropoint]

Hi felixcthecat,

After rethinking this problem I've decided that my original conclusions are based on an invalid line of reasoning. The problem is significantly more complicated. We should not be considering the scaling g with M and r, since g is only an indirect measure of the escape velocity. Instead we should consider the scaling of the escape velocity, so the first thing we should do is to derive it.

Escape velocity can be defined as the minimum speed required for an object to reach infinity. We define gravitational potential energy

E_p = - GM m / r

as the energy required to bring an object from infinity to a radius r. So we can say that the work required to take an object from r to infinity is W = - Ep = GMm/r. We can deliver this work by imparting an initial kinetic energy to the projectile equal to K = 1/2 m v².

The condition for escape velocity is thus

K = W

1/2 m v² = GM m / r

so the expression for the escape velocity is

v = √(2G M / r)

and

v ~ √(M / r).

Therefore increasing both M and r should leave the escape velocity approximately constant.

To see what happens in part 1, we use the same approach I described earlier

v ~ √( (ρ V) / V^1/3)

v ~ ρ^1/2 V^1/3

so the gravity will still increase since square root is more powerful than cube root.

A very convoluted question indeed!

 

[felixcthecat]

The other way around, if you increase to 5M and 2r, gravity increases by a factor of 5/4, which means escape velocity also increases.

oops, yup, rite about that

 

[felixcthecat]

Hi felixcthecat,

After rethinking this problem I've decided that my original conclusions are based on an invalid line of reasoning. The problem is significantly more complicated. We should not be considering the scaling g with M and r, since g is only an indirect measure of the escape velocity. Instead we should consider the scaling of the escape velocity, so the first thing we should do is to derive it.

Escape velocity can be defined as the minimum speed required for an object to reach infinity. We define gravitational potential energy

E_p = - GM m / r

as the energy required to bring an object from infinity to a radius r. So we can say that the work required to take an object from r to infinity is W = - Ep = GMm/r. We can deliver this work by imparting an initial kinetic energy to the projectile equal to K = 1/2 m v².

The condition for escape velocity is thus

K = W

1/2 m v² = GM m / r

so the expression for the escape velocity is

v = √(2G M / r)

and

v ~ √(M / r).

Therefore increasing both M and r should leave the escape velocity approximately constant.

To see what happens in part 1, we use the same approach I described earlier

v ~ √( (ρ V) / V^1/3)

v ~ ρ^1/2 V^1/3

so the gravity will still increase since square root is more powerful than cube root.

A very convoluted question indeed!

i don't get how u got "thus k=w"... is it? ehhe

yea, i was thinking of the escape velocity equation "v = √(2G M / r)" as well.. and atfirst i was wondering whether despite increase in mass and decrease in density (for example 2) whether maybe one is 'more important' than the other hence would influence the escape velocity more despite a same maginitude of change... but from reading the things u deduced, i guess they are all of the same 'importance' and if u change the 2 in the same magnitude in the opposite direction (ie. +/-) then there would be no change of escape velocity.. that's wut i got from ur posts.. is that wut ur thinking? heh, cuz that's wut i'm thinking!!

 

[zeropoint]

i don't get how u got "thus k=w"... is it? ehhe

yea, i was thinking of the escape velocity equation "v = √(2G M / r)" as well.. and atfirst i was wondering whether despite increase in mass and decrease in density (for example 2) whether maybe one is 'more important' than the other hence would influence the escape velocity more despite a same maginitude of change... but from reading the things u deduced, i guess they are all of the same 'importance' and if u change the 2 in the same magnitude in the opposite direction (ie. +/-) then there would be no change of escape velocity.. that's wut i got from ur posts.. is that wut ur thinking? heh, cuz that's wut i'm thinking!!

The work required to reach infinity is W. We perform this work by supplying a kinetic energy equal to W, thus K = W. Is it clear now?

You're thinking is correct, except for the fact that gravity stays the same in part 2 if you change them in the same direction (ie +/+ or -/-), not +/- or -/+.

To summarise:

If you increase density and decrease volume, escape velocity increases since it is more strongly dependent on ρ than V.

If you increase mass and increase radius, escape velocity stays the same since the strength of the positive dependence on M is the same as the strength of the negative dependence on r. Likewise, if you decrease mass and decrease radius, gravity stays the same.

 

[felixcthecat]

The work required to reach infinity is W. We perform this work by supplying a kinetic energy equal to W, thus K = W. Is it clear now?

You're thinking is correct, except for the fact that gravity stays the same in part 2 if you change them in the same direction (ie +/+ or -/-), not +/- or -/+.

To summarise:

If you increase density and decrease volume, escape velocity increases since it is more strongly dependent on ρ than V.

If you increase mass and increase radius, escape velocity stays the same since the strength of the positive dependence on M is the same as the strength of the negative dependence on r. Likewise, if you decrease mass and decrease radius, gravity stays the same.

yayayay!!! thanx zeropoint!! guess there was some complicated equation deducing in the process but i understand now (complicated because of the layout we are confined to..)^^.. was a bit hard to follow through cuz BOS can't really type maths stuff out properly..but i still understand.. heh =D