What is the relationship between tension and mass for an artificial satellite?

[adammish]

Homework Statement

Orbital mechanics

Relevant Equations

$$F=G\frac{mM}{R^2}$$

Hi, I need to prove that the tension for artificial satellite consists of two points of mass m/2 connected by a light rigid rod of length , the tension in the rod is -

$$ T=\frac{3}{4}\frac{Gmm'l}{a^3}-\frac{1}{4}\frac{Gm^2}{l^2} $$

the satellite is placed in a circular orbit of radius a>>l (from the middle point of the rod) around a planet of mass m'. The rod is oriented such that it always points toward the center of the planet. I think that the second term -
$$ -\frac{1}{4}\frac{Gm^2}{l^2} $$
might comes from the mass interaction according to the Newton gravitation law -
$$ F=G\frac{\frac{m}{2}\frac{m}{2}}{l^2} $$
But I don't know how to get the first term (Maybe it related to angular velocity) -
$$ \omega = \frac{(GM)^\frac{1}{2}}{a^\frac{3}{2}} $$
The orbital angular velocity of a planet. Thanks!

 

[sysprog]

That seems to me to be a nice exposition of the problem; please present your attempt at a solution, and then we'll try to help . . .

 

[adammish]

Well, I don't really know how to handle the tension in the rod. In addition, the question didn't provide any information about the tension place (in the middle of the rod? in the end?)
I guess that the additional term gets from the interaction with the planet (because of the m'), and of course from gravity.

I try to do some calculations to get the answer - but I don't think that they are right -

$$ F=\omega ^2R\int dm=\frac{Gm'l}{a^3}\int_{\frac{l}{2}}^{l}mxdx=\frac{Gm'm}{2a^3l}(l^2-\frac{l^2}{4})=\frac{3Gmm'l}{8a^3} $$

 

[sysprog]

Well, I don't really know how to handle the tension in the rod. In addition, the question didn't provide any information about the tension place (in the middle of the rod? in the end?)
I guess that the additional term gets from the interaction with the planet (because of the m'), and of course from gravity.

I try to do some calculations to get the answer - but I don't think that they are right -

A rod is a cylinder (a bent rod is a partial toriod), and you pretty much have to do a triple integral (one for each dimension) -- unless we know the material, we can't really know the tensional characteristics . . .

 

[Delta2]

It is given that the orbit of this satellite is such that the rod points always to the center of the planet. You can draw an important conclusion from this, namely that the tension is along the radius of the orbit, that is along the gravitational force from the planet and also along the gravitational force inbetween the two masses.

Draw FBD for each mass, where you show the Tension, the gravitational force from the planet and the gravitational force from the other mass. All these forces are along the same line, and their algebraic sum is equal to $\frac{m}{2}\omega^2 R$ where $R=a\pm\frac{l}{2}$.

If all go well you will make a system of two equations with two unknowns : The tension $T$ and the angular velocity of the orbit $\omega$. The way this system will be, you can solve for $\omega^2$ relatively easy, and the rest is up to algebra and approximations (given $a>>l$) to solve for the tension $T$.

 

[PeroK]

Well, I don't really know how to handle the tension in the rod. In addition, the question didn't provide any information about the tension place (in the middle of the rod? in the end?)
I guess that the additional term gets from the interaction with the planet (because of the m'), and of course from gravity.

It's not a uniform rod of mass $m$. It's two point masses (each of mass $\frac m 2$), joined by a light rod (of negligible mass).

Note: I tried this assuming that the period was that of Kepler's third law, but that didn't work out. You need to solve for tension and angular frequency as suggested above.

 

[PeroK]

the tension in the rod is -

$$ T=\frac{3}{4}\frac{Gmm'l}{a^3}-\frac{1}{4}\frac{Gm^2}{l^2} $$

I don't see how this answer can be right. You must get a power series in $\frac l a$. I get:
$$T = \frac{3GMml}{4a^3} + \frac{3GMml^3}{16a^5}$$
PS the second term in the book answer is the gravitational attraction between the $\frac m 2$ masses!

You just take my answer to first significant order, neglect the second term and include the self-gravitation of the rod.

PPS which means you can take the period from Kepler's law, as the error is to second order in $l/a$, which is to be neglected.

The problem is, of course, that $M >> m$, so which terms are negligible is not so clear.

 

[adammish]

I don't see how this answer can be right. You must get a power series in $\frac l a$. I get:
$$T = \frac{3GMml}{4a^3} + \frac{3GMml^3}{16a^5}$$
PS the second term in the book answer is the gravitational attraction between the $\frac m 2$ masses!

You just take my answer to first significant order, neglect the second term and include the self-gravitation of the rod.

PPS which means you can take the period from Kepler's law, as the error is to second order in $l/a$, which is to be neglected.

The problem is, of course, that $M >> m$, so which terms are negligible is not so clear.

Can you please explain to me more clearly how did you get the value of T? and what does it mean to get power series of $\frac l a$?

Thank you very much !

 

[PeroK]

Can you please explain to me more clearly how did you get the value of T? and what does it mean to get power series of $\frac l a$?

Thank you very much !

The system has a common angular frequency $\omega$ around the centre of the planet. Using the equation for centripetal force: $$F_c = m\omega^2r$$we see that the outer mass needs a larger force than the inner mass - and the outer mass has a smaller gravitational force. We need tension to increase the force on the outer mass and decrease the force on the inner mass. For the outer mass we have:$$F_c = F_g + T$$ and for the inner mass we have:$$F_c = F_g - T$$This gives you two equations to solve for $\omega$ and $T$.

The quick way is to assume that the period to first significant order in $l/a$ is just the period given by Kepler's law. That gives:$$\omega^2 = \frac{GM}{a^3}$$Then you can solve directly for $T$.

Finally, $T$ can be adjusted down as you have a gravitational attraction between the masses. I would expect that, however, to be several orders of magntide less than the $T$. I actually ran the numbers for a rod of length $1m$ and mass $1kg$ in the ISS and got the gravitational adjustment to be only of the order of $10^{-5}$ of the tension, which was itself only micro-Newtons!

In other words, I think the sensible answer is just the first term and neglect the gravitational force between the masses.

 

[PeroK]

What does it mean to get power series of $\frac l a$?

Let's take the outer mass. The force equation $F_c = F_g + T$ becomes:
$$\frac m 2 \omega^2(a + \frac l 2) = \frac{GMm}{2(a+\frac l 2)^2} + T$$The gravitational term can then be expanded using the binomial theorem (or Taylor series, if you prefer):$$(a + \frac l 2)^{-2} = a^{-2}(1 + \frac l {2a})^{-2} = \frac 1 {a^2}[1 - \frac{l}{a} + \frac{3l^2}{4a^2} - \dots ]$$And for this particular problem, you are effectively asked to neglect terms higher than $l/a$.

 

[adammish]

Now it is all clear!

Thanks a lot!