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elvinc
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Hi,
I have attempted a basic gravitation question from “Classical Mechanics” R. Douglas Gregory (1st ed). I get the answer provided at the back of the book, but it doesn't make sense to me. Can someone help me interpret the answer?
Question: Pg 71, Q3.3
A uniform rod of mass M and length 2a lies along the interval [-a, +a] of the x-axis and a particle of mass m is situated at the point x = x' (*). Find the gravitational force exerted by the rod on the particle.
<< end of first part of question >>
(*) Although not explicitly stated this means the particle is lying on the rod on the x-axis.
Approach:
Using the method from a previous example, split the rod into short segments of length [itex]\delta x[/itex] then applying the inverse square law for universal gravitation the force of attraction between the particle and that small section of the rod is:
[itex]\frac{Gm}{(x^{\prime} - x)^2}.\frac{M\delta x}{2a}[/itex]
Rearranging slightly and summing all these – this becomes an integral
[itex] \frac{GmM}{2a} \int_{-a}^{+a} \! \frac{dx}{(x^ \prime - x)^2} [/itex]
That boils down to
force of attraction = [itex] \frac{GmM} {(x^ \prime)^2 - a^2} [/itex]
which is the answer in the back of the book. Good. Well maybe not.
If I substitute in a couple of values
[itex] x^\prime = 0 [/itex] (the particle is placed at the centre of the rod)
then the expression for the force becomes
[itex] -\frac{GmM} {a^2} [/itex]
a negative force. I don't understand that.
[itex] x^\prime = +a [/itex] (the particle is placed at one end of the rod)
then the expression for the force becomes
[itex] \frac{GmM} {a^2 - a^2} [/itex]
an “infinite” force. I don't understand that.
What is wrong here?
Thanks,
Clive
I have attempted a basic gravitation question from “Classical Mechanics” R. Douglas Gregory (1st ed). I get the answer provided at the back of the book, but it doesn't make sense to me. Can someone help me interpret the answer?
Question: Pg 71, Q3.3
A uniform rod of mass M and length 2a lies along the interval [-a, +a] of the x-axis and a particle of mass m is situated at the point x = x' (*). Find the gravitational force exerted by the rod on the particle.
<< end of first part of question >>
(*) Although not explicitly stated this means the particle is lying on the rod on the x-axis.
Approach:
Using the method from a previous example, split the rod into short segments of length [itex]\delta x[/itex] then applying the inverse square law for universal gravitation the force of attraction between the particle and that small section of the rod is:
[itex]\frac{Gm}{(x^{\prime} - x)^2}.\frac{M\delta x}{2a}[/itex]
Rearranging slightly and summing all these – this becomes an integral
[itex] \frac{GmM}{2a} \int_{-a}^{+a} \! \frac{dx}{(x^ \prime - x)^2} [/itex]
That boils down to
force of attraction = [itex] \frac{GmM} {(x^ \prime)^2 - a^2} [/itex]
which is the answer in the back of the book. Good. Well maybe not.
If I substitute in a couple of values
[itex] x^\prime = 0 [/itex] (the particle is placed at the centre of the rod)
then the expression for the force becomes
[itex] -\frac{GmM} {a^2} [/itex]
a negative force. I don't understand that.
[itex] x^\prime = +a [/itex] (the particle is placed at one end of the rod)
then the expression for the force becomes
[itex] \frac{GmM} {a^2 - a^2} [/itex]
an “infinite” force. I don't understand that.
What is wrong here?
Thanks,
Clive