Finding Planetary Radius using density and escape velocity

In summary: Then ##M## is just the mass of the planet in terms of ##\rho##. OK, so what's ##M##, as I asked in my post? You may represent the density by ##\rho##. Then ##M## is just the mass of the planet in terms of ##\rho##.
  • #1
elDuderino81
7
0
1. "Calculate the radius of a planet with mean density of 3.0x10^3 m2kg-3, from which a golf ball can be thrown to infinity as a velocity of 40 ms-1"


Homework Equations


I've been looking at the equation of:

Vesc=sqroot of 2*G*M/r and rearranging to r=2*G*M/Vesc. However, the trouble is, I'm struggling to get the mass from the density? It appears I don't have enough information, or I'm barking up the wrong tree so to speak?


The Attempt at a Solution



can anyone point me in the right direction please?
 
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  • #2
elDuderino81 said:
1. "Calculate the radius of a planet with mean density of 3.0x10^3 m2kg-3, from which a golf ball can be thrown to infinity as a velocity of 40 ms-1"

Homework Equations


I've been looking at the equation of:
fra
Vesc=sqroot of 2*G*M/r and rearranging to r=2*G*M/Vesc.


Recheck your algebra there. What happened to the square root?

However, the trouble is, I'm struggling to get the mass from the density? It appears I don't have enough information, or I'm barking up the wrong tree so to speak?

The Attempt at a Solution



can anyone point me in the right direction please?

Start with the conservation of energy statement ##\frac{1}{2}mv_e^2 = \frac{GMm}{r}## that leads to the equation you started with. ##m## is the mass of the golf ball (cancels out), while ##M## is the mass of the planet.

Now find an expression for ##M## in terms of the density and the radius. Assume the planet is a spherical ball of radius ##r##. What's the enclosed volume of a perfect sphere?
 
  • #3
Curious3141 said:
Recheck your algebra there. What happened to the square root?
Start with the conservation of energy statement ##\frac{1}{2}mv_e^2 = \frac{GMm}{r}## that leads to the equation you started with. ##m## is the mass of the golf ball (cancels out), while ##M## is the mass of the planet.

Now find an expression for ##M## in terms of the density and the radius. Assume the planet is a spherical ball of radius ##r##. What's the enclosed volume of a perfect sphere?

The enclosed volume of a perfect sphere is is V=(3/4)∏*r3, and when rearranging the previous equation I get r=GM/0.5Ve^2 and M=G/0.5V^2*r?

I'm still struggling to see what I can do with this, as it appears that to find r I need M and to find M I need r? I'm really confused :-(
 
Last edited:
  • #4
elDuderino81 said:
The enclosed volume of a perfect sphere is is V=(3/4)∏*r3, and when rearranging the previous equation I get r=GM/0.5Ve^2 and M=G/0.5V^2*r?

I'm still struggling to see what I can do with this, as it appears that to find r I need M and to find M I need r? I'm really confused :-(

You have ##V##. What's the relationship between mass, density and volume? Hence what is ##M## in terms of ##r##?

Replace ##M## with that expression. Rearrange to isolate ##r## on one side of the equation. That's just simple algebra. But be careful with it - you seem prone to making mistakes with this. The expressions you wrote are ambiguous (you should use LaTex formatting), but there seems to be mistake with the rearrangement here too.
 
  • #5
elDuderino81 said:
The enclosed volume of a perfect sphere is is V=(3/4)∏*r3, ... I'm really confused :-(

Yes, you are confused.

Is your planet a flabby sphere? You should recheck your formula for the volume of a sphere. I also didn't understand the units of average density in the OP for the planet. The units of density are ML^-3.
 
  • #6
SteamKing said:
Yes, you are confused.

Is your planet a flabby sphere? You should recheck your formula for the volume of a sphere. I also didn't understand the units of average density in the OP for the planet. The units of density are ML^-3.

Hi, sorry about the typo, i should have wrote (4/3)∏r^3

In regards to units of density, again that was a typo and should read 3*10^3 kg M^-3, which is what has been provided in the problem set.
 
  • #7
elDuderino81 said:
Hi, sorry about the typo, i should have wrote (4/3)∏r^3

In regards to units of density, again that was a typo and should read 3*10^3 kg M^-3, which is what has been provided in the problem set.

OK, so what's ##M##, as I asked in my post? You may represent the density by ##\rho##.
 

Related to Finding Planetary Radius using density and escape velocity

1. How can I determine the planetary radius using density and escape velocity?

The planetary radius can be calculated using the formula: R = (3M/4πρ)^1/3 * (2GM/v)^2/3, where R is the radius, M is the mass, ρ is the density, G is the gravitational constant, and v is the escape velocity.

2. What is the relationship between density and escape velocity?

The density of a planet is directly proportional to its escape velocity. This means that the higher the density, the higher the escape velocity required for an object to leave the planet's gravitational pull.

3. Can I use the same formula for all planets?

No, the formula for calculating planetary radius using density and escape velocity can only be used for rocky planets with a solid surface. Gas giants and other types of planets have different formulas for determining their radius.

4. How accurate is this method of finding planetary radius?

This method can provide a reasonably accurate estimate of a planet's radius, but it is not exact. Other factors such as the planet's composition and atmospheric conditions can affect the results.

5. Is there a simpler way to find a planet's radius?

Yes, there are other methods for determining a planet's radius, such as using telescopic observations or analyzing the planet's shadow during a transit. However, these methods may not be feasible for all planets and may require more advanced equipment.

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