1. ## Fitzpatrick mechanics question

9. A body is fired vertically from the surface of the moon with initial velocity v0.
(i) Find the velocity of the body as a function of its distance r from the centre of the moon.
(ii) Find the escape velocity. (Take the radius of the moon as R and gravitational acceleration on the moon as g/6.)

(i) v^2=v0^2+gR/3 (R/r - 1)
(ii) sqrt (gR/3)

2. ## Re: Fitzpatrick mechanics question

Hey.

Just a couple of things. Firstly I think it's easiest to visualize (by drawing a diagram with the moon/centre/radiis etc) what's going on so you can intuitively work out stuff. Secondly you need to assume that the acceleration due to gravity at the Moon's surface is $\frac{g}{6}$ for both parts of the question. Thirdly you need to assume that the acceleration due to gravity at a point outside the Moon is inversely proportional to the square of the distance from the centre of the Moon i.e. $\ddot{x}\propto-\frac{1}{x^{2}}\hspace{0.5cm}->\ddot{x}=-\frac{k}{x^{2}}$

$\\\ddot{x}=\frac{-k}{x^{2}}\\At the surface of the Earth\hspace{0.5cm}\ddot{x}=\frac{-g}{6},x=R.\\Thus k=\frac{R^{2}g}{6}\\v\frac{dv}{dx}=\frac{-R^{2}g}{6x^{2}}\\\int^{v}_{v_0}vdv=\frac{-r^{2}g}{6}\int^{r}_{R}x^{-2}dx\\\frac{1}{2}[v^{2}]^{v}_{v_0}=\frac{R^{2}g}{6}[\frac{1}{x}]^{r}_{R}\\\frac{v^{2}-{v_0}^{2}}{2}=\frac{R^{2}}{6r}-\frac{Rg}{6}\\v^{2}={v_0}^{2}+\frac{R^{2}g}{3r}-\frac{Rg}{3}\\={v_0}^{2}+\frac{Rg}{3}(\frac{R}{r}-1)\\For v_e, x>Rfor the body to leave the moon\\\frac{{v_e}^{2}}{2}<\frac{R^{2}g}{6R}\\{v_e}^{2}>\frac{Rg}{3}\\v_e>\sqrt{\frac{Rg}{3}}$

Sorry don't know how to put space in the text, so here's a link to a picture of it mechanics q.png

3. ## Re: Fitzpatrick mechanics question

Originally Posted by BenHowe
Sorry don't know how to put space in the text, so here's a link to a picture of it mechanics q.png
Use \text{}. eg: \text{This is a phrase} would produce:
$\text{This is a phrase}$

4. ## Re: Fitzpatrick mechanics question

Originally Posted by envlam
Use \text{}. eg: \text{This is a phrase} would produce:
$\text{This is a phrase}$
$\text{omg that is so much better}$. I was just going \hspace{0.5cm} but then my equation become too long or I fked something up and it became invalid -_-

5. ## Re: Fitzpatrick mechanics question

Thank you for your help. I still have a few questions though. isn't the acceleration due to gravity g/6 at all points around the moon. Also, when you say at the surface of the earth, do you mean on the moon? and I don't quite understand what you did for part (ii)

6. ## Re: Fitzpatrick mechanics question

Ok, well firstly the acceleration due to gravity is only $\frac{g}{6}$ on the surface of the moon. The easiest way to think about this is, consider the Earth. Is g constant? No. It varies with altitude,location etc. This is why I put in that statement about the acceleration, that allows you to determine the acceleration for the body at all positions not on the surface of the moon i.e. some distance. They just include that stuff so you can find the constant term. You also have to be careful that the acceleration is always negative because of how you have defined distance with respect to the centre of the Moon's mass or Moon.

To answer your next point, when I said Earth I meant moon, that was a complete fk up

Ok and for part (ii) I skipped a few steps because I was afraid my equation would be too long to pop it in one post. Anyways, all I did was get the equation I derived in (i) but without putting in the limits (think of the integral in (i) as indefinite not definite). Then this is where the logic comes into play. Escape velocity is the minimum velocity required for the body to JUST escape the gravitational force of the Moon and enter space. Hence you assign the condition $x>R$, so you will be able to find the minimum velocity for the body to escape! Also technically you should specify direction i.e. away from the Moon's centre of mass etc. So the answer they've given is technically not correct it should be > not equal to it, because when it is equal it will orbit, but that's bordering into the realm of physics.

Here, let me post a picture of my diagram and also of my working.

mechanic q diagram attempt 2.JPG

$\frac{vdv}{dx}=\frac{-R^{2}g}{6x^{2}}\\vdv=\frac{-R^{2}g}{6}x^{-2}dx\\\text{Integrate the left hand side with respect to v and the right hand side with respect to x}\\\int vdv=\frac{-R^{2}g}{6}\int x^{-2}dx\\\frac{v^{2}}{2}=\frac{R^{2}g}{6x}\hspace{0.5 cm}\text{ignore the constant since their final answer doesn't have two terms and you'll also find that if you do the question with}\\\text{the constant term you will just get}\hspace{0.5cm}v_e>v_0\\\text{Make x the subject of the equation i.e.}\hspace{0.5cm} x=\frac{gR^{2}}{3v^{2}}\\x>R\hspace{0.5cm}\text{in order for the body to escape the Moon}\\\text{So use this condition i.e.}\hspace{0.5cm} R>\frac{gR^{2}}{3v^{2}}\\3v^{2}>gR\\v^{2}>\frac{gR}{3}\\v>\sqrt{\frac{gR}{3}}\hspace{0.5cm}\text{away from the centre of the Moon}$

Hope this helps

7. ## Re: Fitzpatrick mechanics question

Originally Posted by BenHowe
Ok, well firstly the acceleration due to gravity is only $\frac{g}{6}$ on the surface of the moon. The easiest way to think about this is, consider the Earth. Is g constant? No. It varies with altitude,location etc. This is why I put in that statement about the acceleration, that allows you to determine the acceleration for the body at all positions not on the surface of the moon i.e. some distance. They just include that stuff so you can find the constant term. You also have to be careful that the acceleration is always negative because of how you have defined distance with respect to the centre of the Moon's mass or Moon.

To answer your next point, when I said Earth I meant moon, that was a complete fk up

Ok and for part (ii) I skipped a few steps because I was afraid my equation would be too long to pop it in one post. Anyways, all I did was get the equation I derived in (i) but without putting in the limits (think of the integral in (i) as indefinite not definite). Then this is where the logic comes into play. Escape velocity is the minimum velocity required for the body to JUST escape the gravitational force of the Moon and enter space. Hence you assign the condition $x>R$, so you will be able to find the minimum velocity for the body to escape! Also technically you should specify direction i.e. away from the Moon's centre of mass etc. So the answer they've given is technically not correct it should be > not equal to it, because when it is equal it will orbit, but that's bordering into the realm of physics.

Here, let me post a picture of my diagram and also of my working.

mechanic q diagram attempt 2.JPG

$\frac{vdv}{dx}=\frac{-R^{2}g}{6x^{2}}\\vdv=\frac{-R^{2}g}{6}x^{-2}dx\\\text{Integrate the left hand side with respect to v and the right hand side with respect to x}\\\int vdv=\frac{-R^{2}g}{6}\int x^{-2}dx\\\frac{v^{2}}{2}=\frac{R^{2}g}{6x}\hspace{0.5 cm}\text{ignore the constant since their final answer doesn't have two terms and you'll also find that if you do the question with}\\\text{the constant term you will just get}\hspace{0.5cm}v_e>v_0\\\text{Make x the subject of the equation i.e.}\hspace{0.5cm} x=\frac{gR^{2}}{3v^{2}}\\x>R\hspace{0.5cm}\text{in order for the body to escape the Moon}\\\text{So use this condition i.e.}\hspace{0.5cm} R>\frac{gR^{2}}{3v^{2}}\\3v^{2}>gR\\v^{2}>\frac{gR}{3}\\v>\sqrt{\frac{gR}{3}}\hspace{0.5cm}\text{away from the centre of the Moon}$

Hope this helps
Thanks for your detailed response. I'm just confused about the last bit. why is it x>R? wouldn't this be just above the moon's surface, not escaping its gravitational field. I thought escape velocity was found by putting x approaches infinity.

8. ## Re: Fitzpatrick mechanics question

no no. You're getting it confused with that physics formula I think where they derive Ep and use the defintion to account for the negative in the equation. For mx2 escape velocity is the velocity required for a body to leave the planet, dont go into gravitational field in e2.

9. ## Re: Fitzpatrick mechanics question

Originally Posted by BenHowe
no no. You're getting it confused with that physics formula I think where they derive Ep and use the defintion to account for the negative in the equation. For mx2 escape velocity is the velocity required for a body to leave the planet, dont go into gravitational field in e2.
It's still the speed required to escape the gravitational field. Any (positive) launch speed would make it leave the surface of the planet of course (e.g. just imagine throwing a ball up).

10. ## Re: Fitzpatrick mechanics question

*Facepalm

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•