Variation of escape velocity with equal surface gravity

[sachin]

Homework Statement

the question says "There are two Planets, each with the same surface gravity, but Planet 1 has a greater radius and is less massive then Planet 2. Which of these planets has a greater escape velocity? "

Relevant Equations

__ GMm/R + 1/2 m v2 = 0, so v escape = root 2GM/R, which is not directly dependent on g, as g = GM/R2

the question says "There are two Planets, each with the same surface gravity, but Planet 1 has a greater radius and is less massive then Planet 2. Which of these planets has a greater escape velocity? "
(A) Planet 1
(B) Planet 2
(C) Both have the same escape velocity because surface gravity is equal
(D) Not enough information is given

we have,
__ GMm/R + 1/2 m v2 = 0, so v escape = root 2GM/R, which is not directly dependent on g, as g = GM/R2,

my concern is probably the question given in ap physics worksheet in the link given below is needing correction as if mass of the planet M is less and also it has more radius than the other planet, it can never have the same surface gravity which is given by g = GM/R2,

http://content.njctl.org/courses/sc...-gravitation-multiple-choice-2-2015-12-06.pdf

the answer given in the sheet is option is option B i.e Planet 2,

how is it possible,
thanks.

 

[hutchphd]

The answer is more or less given to you. They are spheres and M1 has less mass.

 

[Ibix]

if mass of the planet M is less and also it has more radius than the other planet, it can never have the same surface gravity which is given by g = GM/R2,

I agree - I don't think the question is consistent. If the surface gravity is to be the same and the mass is less, the radius must also be less.

 

[Orodruin]

v escape = root 2GM/R, which is not directly dependent on g, as g = GM/R2

... but you know that $g$ is the same for both planets so you can use that to eliminate either $M$ or $R$. You know that $M = gR^2/G$ and so
$$
v_{\rm esc} = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2GgR^2}{RG}} = \sqrt{2gR}
$$
Draw your conclusions from there.

However:

Planet 1 has a greater radius and is less massive then Planet 2.

This is impossible. If they are supposed to have the same $g$, then $M/R^2$ must be the same for both planets. If the radius increases, then so must the mass in order to maintain the same $g$ at the surface.

The answer is more or less given to you. They are spheres and M1 has less mass.

Having less mass does not necessarily imply lower escape velocity, it depends on the radius. Compare the Sun to a (hypothetical) primordial black hole of an Earth mass or so.

 

[Orodruin]

It should be noted that $M \propto \rho R^3$ where $\rho$ is the average density. This would make $g \propto \rho R$ and therefore the density must be lower if the radius increases. This could possibly be what the questioner intended, but generally I would assume "more massive" to refer to the actual mass and not the specific mass (i.e., density) of an object.

 

[sachin]

so, hope i can conclude that the given question is wrong, thanks.

 

[Orodruin]

Wrong in the sense of the stated premise not being correct, yes. It would be a reasonable question if they removed either the statement about the radius or that of the mass.

 

[haruspex]

Wrong in the sense of the stated premise not being correct, yes. It would be a reasonable question if they removed either the statement about the radius or that of the mass.

Clearly the answer is E: too much information is given