Although you don't need to for this part of the question, it's best to draw a diagram to help you understand the question better.
(Note I'm using x'' as the acceleration of the object. )
Now x''=-k/x<SUP>3</SUP>
At the surface of the earth, x''= -g , and x = R
.: -g = -k/R<SUP>3</SUP>
k = gR<SUP>3 </SUP>as required.
I know you didn't ask for the second bit, but I've done it anyway and I thought I'd post it up in case you, or anyone else, was unsure.
(ii) Show that v, the velocity of the particle, is given by
v<SUP>2</SUP> = gR<SUP>3</SUP>/x<SUP>2 </SUP>- (gR - u<SUP>2</SUP>)
Now x'' = -k/x<SUP>3</SUP>
and k = gR<SUP>3</SUP> from (i)
Then x''= -gR<SUP>3</SUP>/x<SUP>3</SUP>
Using x'' = d/dx (1/2v<SUP>2</SUP>)
d/dx(1/2v<SUP>2</SUP>) = -gR<SUP>3</SUP>/x<SUP>3</SUP>
Intergrating both sides w.r.t. x
1/2v<SUP>2</SUP> = -gR<SUP>3</SUP> ∫x<SUP>-3</SUP>dx
v<SUP>2</SUP> = -2gR<SUP>3</SUP> (x<SUP>-2</SUP>/-2) + C
v<SUP>2</SUP> = gR<SUP>3</SUP>/x<SUP>2</SUP> + C
At x = R, v = u
u<SUP>2</SUP> = gR<SUP>3</SUP>/R<SUP>2</SUP> + C
C = u<SUP>2</SUP> - gR
v<SUP>2</SUP> = gR<SUP>3</SUP>/x<SUP>2</SUP> + u<SUP>2</SUP> - gR
v<SUP>2</SUP> = gR<SUP>3</SUP>/x<SUP>2</SUP> - (gR - u<SUP>2</SUP>) as required
(iii) It can be shown that
x = √(R<SUP>2</SUP> + 2uRt - (gR - u<SUP>2</SUP>)t<SUP>2</SUP>)
(DO NOT PROVE THIS)
Show that if u ≥ √(gR) then the particle will not return to the planet.
My answer to this wasn't as rigorous as I'd have liked it to be. Hopefully someone else can post their one up as well.
v<SUP>2</SUP> = gR<SUP>3</SUP>/x<SUP>2</SUP> - (gR - u<SUP>2</SUP>) from (ii)
Now v = 0 for max height x = H
0 = gR<SUP>3</SUP>/H<SUP>2</SUP> - gR + u<SUP>2</SUP>
gR - u<SUP>2</SUP> = gR<SUP>3</SUP>/H<SUP>2</SUP>
H<SUP>2</SUP> = gR<SUP>3</SUP>/(gR - u<SUP>2</SUP>)
H = √(gR<SUP>3</SUP>/(gR - u<SUP>2</SUP>))
As (gR - u<SUP>2</SUP>) --> 0
H --> ∞
gR - u<SUP>2</SUP> ≤ 0 This is the line I'm not sure of. It gives the required answer, but it seems to give an undefined value of H.
u<SUP>2</SUP> ≥ gR
u ≥ √(gR) as required.
(iv) If u < √(gR) then the particle reaches a point whose distance from the centre of the planet is D, and then falls back.
(1) Use the formula in part (ii) to find D in terms of u, R and g.
(2) Use the formula in part (iii) to find the time taken for the particle to return to the surface of the planet in terms of u, R and g.
(1)
v<SUP>2</SUP> = gR<SUP>3</SUP>/x<SUP>2</SUP> - (gR - u<SUP>2</SUP>)
v = 0 for max height x = D
0 = gR<SUP>3</SUP>/D<SUP>2</SUP> - gR + u<SUP>2</SUP>
gR - u<SUP>2</SUP> = gR<SUP>3</SUP>/D<SUP>2</SUP>
D<SUP>2</SUP> = gR<SUP>3</SUP>/(gR - u<SUP>2</SUP>)
D = √(gR<SUP>3</SUP>/(gR - u<SUP>2</SUP>))
<SUP></SUP>
<SUP></SUP>
(2) x = √(R<SUP>2</SUP> 2uRt - (gR - u<SUP>2</SUP>)t<SUP>2</SUP>)
Particle returns to the planet when x = R
R = √(R<SUP>2</SUP> 2uRt - (gR - u<SUP>2</SUP>)t<SUP>2</SUP>)
R<SUP>2</SUP> = R<SUP>2</SUP> 2uRt - (gR - u<SUP>2</SUP>)t<SUP>2</SUP>
gRt<SUP>2</SUP> - u<SUP>2</SUP>t<SUP>2</SUP> - 2uRt = 0
t [ (gR - u<SUP>2</SUP>) t - 2uR] = 0
t = 0 (trivial) or (gR - u<SUP>2</SUP>) t - 2uR = 0
(gR - u<SUP>2</SUP>) t = 2uR
t = 2uR/(gR - U<SUP>2</SUP>)
If anyone sees something wrong with anything in there just let me know and I'll try and fix it. Also, if anyone knows how to do (iii) would they be able to post it up and explain.
