Meteor Entering Orbit of a Sun (cons. energy & momentum) [very hard algebra]

AI Thread Summary
The discussion focuses on solving a physics problem involving a meteor's trajectory influenced by the sun's gravitational force. The user attempts to derive the distance of closest approach, denoted as b, using conservation of momentum and energy equations. Initial and final energy equations are set up, but the user struggles with algebraic manipulation to isolate b. After several attempts and corrections, a key mistake is identified in the quadratic formula application, leading to a successful resolution of the problem. The conversation highlights the importance of careful algebra in physics problem-solving.
DD31
Messages
10
Reaction score
0

Homework Statement


Alright, so I've given this guy a few cracks, and I think I'm close to it, but apparently I'm not right, so here's how it goes. Simple in concept.

A meteor moves toward the solar system with speed v0 in a direction such that it would miss the sun by a distance d if it were not attracted by the sun's gravitational force. Denote the mass of the sun by M. Find the distance b of the meteor from the sun at the point of closest approach in terms of v0, d, and M (and G, the gravitational constant).

http://img515.imageshack.us/img515/8376/lg0704figgx8.jpg
http://g.imageshack.us/img515/lg0704figgx8.jpg/1/


Homework Equations


They effectively give me a few formulas in the helper questions provided:

The momentum and energy are conserved
Initial energy = 1/2mv0^2
Final energy = 1/2mV^2 - GMm/b

Initial Momentum = mv0d
Final Momentum = mVb

Where V=the final velocity of the object at point b

The Attempt at a Solution



I feel like the error here is most likely in my algebra, as I'm having a ton of troble easily isolating b. But here are the ways I've tried so far:

I always start with the equations:

1/2mv0^2 = 1/2mV^2 - GMm/b

and

mv0d = mVb.

I then cancel out all the m's and multiply out the 2 in the top one to get:

v0^2 = V^2 - 2GM/b

and

v0d = Vb

From this point, I'll either solve for b in the top or solve for V in the bottom. If I solve for b in the top, I get:

V^2 - v0^2 = 2GM/b, b = 2GM/(V^2 - v0^2)

Then I would solve for V in the bottom one:

V = v0d/b

Subbing that in gets

b = 2GM/((v0d/b)^2 - v0^2)

And this is where I think I go wrong...trying to get b out of this. I've tried rearranging the equation to bring all the b's to one side, I've tried resubstituting stuff...most recently I tried quad formula:

I factored out the v0's on the bottom:

2GM/((v0^2)(d^2/b^2-1)) = b

b(d^2/b^2 - 1) = 2GM/v0^2

d^2/b - b = 2GM/v0^2

d^2 - b^2 = 2GMb/v0^2

b^2 + 2GMb/(v0^2) - d^2 = 0

then, using 1 as (a), 2GM/v0^2 as (b) and -d^2 as (c), I did quad form and got:

(-2GM/(v0^2)+sqrt((2GM/v0^2)^2-4d^2))/2

or

http://img395.imageshack.us/img395/482/e5a7daf972eacf000cabb54ld4.gif
http://g.imageshack.us/img395/e5a7daf972eacf000cabb54ld4.gif/1/


But that was not correct


I'm officially stumped by this point...I've taken 4 cracks at it, to no avail. Am I setting up the formulas incorrectly, or am I just screwing up on the algebra over and over?
 
Last edited by a moderator:
Physics news on Phys.org
You are really close.

What is "-4ac", when c is -d2? (And a=1 of course)
 
haha...wow. I see it...it just should be + 4d^2. Just put that in and it worked. Thanks a ton...Almost sure I wouldn't have picked that out without doing the whole thing over again.

Thanks.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top