Why do component vectors not work ?

In summary, the conversation discusses different approaches to solving a physics problem involving the net gravitational force exerted on a spaceship by two asteroids. One person suggests breaking up the asteroids into smaller components and placing them along the axes, but this method is found to be incorrect because it does not account for the angle at which the asteroids are relative to the spaceship. The correct method involves calculating the components of the force using the asteroid's mass and the angle, and then adding them together. The conversation ends with a suggestion to review vector addition and practice exercises to gain a better understanding.
  • #1
Newton22
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1. Homework Statement

The millennium eagle passes between a pair of asteroids, as shown. If the mass of the spaceship is 2.50 x 10⁷kg and the mass of each asteroid is 3.50 x 10¹¹kg, find the net gravitational force exerted on the millennium eagle at point (A) and point (B). Treat the asteroids and the spaceship as if they were point objects

Homework Equations



F = G(m₁m₂)/r²

The Attempt at a Solution


This is not really a question I need help answering as the solution is laid out by the author. My question is more about the fundamentals of doing this type of problem.

As you can see in part (A) found the value of R using the Pythagorean theorm and then added up the components to find the net force acting.

I wanted to try and solve it a different way. So I decided to fist calculate the components and then just add them up. This seemed easier and more straightforward to me as it requires less steps.

F1 = F1x + F1y
F2 = F2x + F2y

When calculating the F1x and F1y I got ridiculously different answers than what F1x and F1y is in the book.

Using the universal gravitational law I got

F1x = (6.67x10^-11)(2.50x10^7)(3.50x10^11)/(3000^2)N = 64.8N
As you can see this Is not what I am supposed to get for F1x..

Its seems like my strategy will not work. Is there a reason why? Can somebody please explain this? Thanks!
 
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  • #2
Your answer is wrong because you basically shifted the whole asteroid to point B.
If you want to calculate it like you did, you'd have to break up the given asteroid accordingly into two pieces with smaller masses, and place them along the axes.

Imagine this, you are being hosed with a water pipe. Now you say I want to take two components of this flow of water, and going by the way you did it, you place two hoses along the axes hitting you. That's not two smaller components of a hose, that's the flow of two entire hoses!
 
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  • #3
Mickey Tee said:
Your answer is wrong because you basically shifted the whole asteroid to point B.
If you want to calculate it like you did, you'd have to break up the given asteroid accordingly into two pieces with smaller masses, and place them along the axes.

Imagine this, you are being hosed with a water pipe. Now you say I want to take two components of this flow of water, and going by the way you did it, you place two hoses along the axes hitting you. That's not two smaller components of a hose, that's the flow of two entire hoses!

So , essentially what I was trying to do was to take two asteroids of the same mass , one 3.00km horizontally from the ship and one 1.50km vertically from the ship and try to equate that to one asteroid of that mass that is 3.00km horizontally and 1.50km vertically from it .. which now makes sense to me why it wouldn't work, because that would be double the mass! Right?

Since we are treating the steroids as point particles, would it be correct to assume that half the mass causes gravity in the x direction and half of it in the y direction?
 
  • #4
No it depends on the angle at which the Asteroid is relative to the spaceship. The masses would be msin26.6 and mcos26.6 repectively. And that hose analogy wasn't entirely accurate, you'd feel 1.414 times the force of each hose if they were placed at equal distances from you, i.e at an angle of 45 degrees.
 
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  • #5
Mickey Tee said:
No it depends on the angle at which the Asteroid is relative to the spaceship. The masses would be msin26.6 and mcos26.6 repectively. And that hose analogy wasn't entirely accurate, you'd feel 1.414 times the force of each hose if they were placed at equal distances from you, i.e at an angle of 45 degrees.

Thank you. So can you explain how it is the masses are dependent on the angle exactly?
Here's what I think.

The force of gravity itself is along the line that connects the two particles. This force depends on the masses of the particles as well as the inverse square of total distance between them. The force itself is linear and the components (x,y) do not actually exist. But can still be computed if the line of action is made to be at some angle relative to some axis.

I'm stuck man help me.. I got nowhere with that .. how can this be explained ??
 
  • #6
I think you should go over vector addition and do some exercises. It'll make sense, everything is already explained.
 
  • #7
Mickey Tee said:
I think you should go over vector addition and do some exercises. It'll make sense, everything is already explained.
Yeah but mass is a scalar.. and your saying the mass would be distributed in the x and y directions based on the angle that it makes with the axis... I'm trying to put it in perspective to understand why it's true
 
  • #8
I said the same result would be produced if you placed two masses of msinθ and mcosθ along the axes at ry and rx distance away.
Sorry if my answer was confusing.
 
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1. Why do component vectors not work in some situations?

Component vectors do not work in some situations because they are only applicable to linear systems. Non-linear systems, such as those with complex or chaotic behaviors, cannot be accurately represented using component vectors.

2. Can component vectors be used to represent all types of vectors?

No, component vectors can only be used to represent vectors in two or three dimensions. They cannot be used to represent higher-dimensional vectors, such as those used in abstract algebra or linear algebra.

3. How do component vectors differ from other vector representations?

Component vectors differ from other vector representations, such as geometric vectors or polar vectors, in that they break down a vector into its individual components along specific axes. This allows for easier calculations and analysis of the vector's properties.

4. Are component vectors always orthogonal?

No, component vectors are not always orthogonal. They are only orthogonal when the axes used to represent the vector are also orthogonal, such as in the x-y plane. In other cases, the component vectors may have different angles between them.

5. How accurate are component vectors in representing real-world phenomena?

Component vectors can be highly accurate in representing real-world phenomena, as long as the system being modeled is linear and the axes chosen are appropriate for the situation. However, as mentioned before, they may not accurately represent non-linear systems or higher-dimensional vectors.

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