Minimum velocity to throw a object to another planet

[Yalanhar]

Homework Statement

Two planets A and B (mass m for both) are separated by a distance of D. A object is thrown from the surface of planet A towards planet B. Radius of planet A and B are R and 2R. Find the minimum velocity to throw the object from planet A to B.

Relevant Equations

$dW=\frac {GMm}{r^2}dr$
$W=\Delta U_p$
$K_o +U_o = K+U$

So I integrated the work done on the object by both planets. Work1 is until x, and Work2 is from x to d. Where x is the point where both gravitational forces are equal.

$W_1=\int_0^x \frac{GMm}{r^2}dr - \int_0^x \frac{GMm}{(3R+D-r)^2}dr$
$W_2=\int_x^D \frac{GMm}{(3R+D-r)^2}dr - \int_x^D \frac{GMm}{r^2}dr$

$W_t=W_1+W_2$

$W_t = \Delta U_p$

Chosing zero at A's surface

$W_t=U$

And then I have to conserve the energy and calculate $v_o$. However, the algebraic is a pain and I still have that $x=\frac{3R+D}{2}$.

Any help? The answer is $v_o = \sqrt \frac{GM(D-R)}{R(3R+D)}$

 

[Orodruin]

The potential energy is always due to both bodies. You cannot leave the force of 2 out of the journey from 1 to x and you cannot leave the force of 1 out of the journey from x to D.

Instead of looking at the forces, I suggest looking at the potentials, which also add linearly. Also note that the potential difference between the planets is going to be zero due to symmetry so the total work done when the object arrives at planet 2 is always going to be zero.

 

[jbriggs444]

Also note that the potential difference between the planets is going to be zero due to symmetry so the total work done when the object arrives at planet 2 is always going to be zero.

The two planets have the same masses but differing radii. A surface-to-surface trajectory will not be symmetric.

So I integrated the work done on the object by both planets. Work1 is until x, and Work2 is from x to d. Where x is the point where both gravitational forces are equal.

An integral from r=0 (at the center of planet A) to r=x (at the point of maximum gravitational potential between A and B) does not seem very relevant.

What is the minimum feasible velocity (or energy) of the thrown object when it reaches position x? Using that information, how could you go about determining the velocity (or energy) with which it must have been thrown?

 

[Yalanhar]

What is the minimum feasible velocity (or energy) of the thrown object when it reaches position x? Using that information, how could you go about determining the velocity (or energy) with which it must have been thrown?

That seems easier, then I integrate until r=x?

 

[Yalanhar]

Actually, the integral goes from R to x. 0 would get a divergent integral.

 

[PeroK]

Actually, the integral goes from R to x. 0 would get a divergent integral.

I can't help wondering about this problem. Is $D$ much greater than $R$, as you might expect? Or, if $D$ is only a bit bigger than $3R$, does that make any difference?

In the second case, I wonder what happens to the object when it lands on the surface of planet B?

Hmm!

 

[Yalanhar]

I can't help wondering about this problem. Is $D$ much greater than $R$, as you might expect? Or, if $D$ is only a bit bigger than $3R$, does that make any difference?

In the second case, I wonder what happens to the object when it lands on the surface of planet B?

Hmm!

It doesn't say anything about it. Just that the distance between them is D. I tried solving the first integral but ended up with natural logarithms, so I think there is something wrong but I can't understand what.

 

[PeroK]

It doesn't say anything about it. Just that the distance between them is D. I tried solving the first integral but ended up with natural logarithms, so I think there is something wrong but I can't understand what.

In general, never integrate unless you have to. Can you describe the motion of a projectile fired between two planets?

Assume $D$ is large, to begin with at least.

 

[DEvens]

In general, never integrate unless you have to. Can you describe the motion of a projectile fired between two planets?

Assume $D$ is large, to begin with at least.

The advice about not integrating is very good advice. In this case, unless my guess is wrong, the only integral you need is to determine the difference in potentials. And that integral ought to already be done.

If D is not large compared to R then it's a very different problem. For example, if you wait a while the answer may be zero.

 

[Yalanhar]

The advice about not integrating is very good advice. In this case, unless my guess is wrong, the only integral you need is to determine the difference in potentials. And that integral ought to already be done.

If D is not large compared to R then it's a very different problem. For example, if you wait a while the answer may be zero.

The inegral I'm trying is for the potenial energy. I'll try wih $D>>R$, the list of the exercise is a hard one but not that hard

 

[PeroK]

The inegral I'm trying is for the potenial energy. I'll try wih $D>>R$, the list of the exercise is a hard one but not that hard

Physics involves applying mathematics. Until you understand a problem you can't really know what mathematics to apply.

The physics involves describing and understanding the problem: in this case the motion of a projectile fired between distant planets.

Once you have understood the physics, then you know what mathematics to apply.

In this case knowing that the solution has something to do with gravitational forces and potentials is not enough.

 

[Yalanhar]

Physics involves applying mathematics. Until you understand a problem you can't really know what mathematics to apply.

The physics involves describing and understanding the problem: in this case the motion of a projectile fired between distant planets.

Once you have understood the physics, then you know what mathematics to apply.

In this case knowing that the solution has something to do with gravitational forces and potentials is not enough.

I understood that the object is decelerating until x. Then the planet B exerces a greater force so is now accelerating. But the accelerating does not increase linearly. Thats why I used work and potential energy... I thought it would be like the escape velocity calculus...

 

[jbriggs444]

I understood that the object is decelerating until x. Then the planet B exerces a greater force so is now accelerating. But the accelerating does not increase linearly. Thats why I used work and potential energy... I thought it would be like the escape velocity calculus...

You could integrate work. Or you could use potential energy. Either way will get you to a solution.

But potential energy is already the path integral of incremental work. Can you articulate why you are integrating twice? [And thereby integrating total work rather than incremental work]

 

[Yalanhar]

You could integrate work. Or you could use potential energy. Either way will get you to a solution.

But potential energy is already the path integral of incremental work. Can you articulate why you are integrating twice?

But for I don't know how to calculate the potential energy of a object by two planets. The formula:
$U=-\frac{GMm}{r}$ Isn't only for 1 planet? So i am integrating work to calculate the potential energy by using $W_t = \Delta U_p$. Work of a conservative force only changes the potential energy.
Then I would use
$K_o + U_o = K + U$
$\frac {1}{2} mv^2 - U = 0$

 

[jbriggs444]

But for I don't know how to calculate the potential energy of a object by two planets. The formula:
$U=-\frac{GMm}{r}$ Isn't only for 1 planet? So i am integrating work to calculate the potential energy by using $W_t = \Delta U_p$. Work of a conservative force only changes the potential energy.

Huh? What?

Someone else integrated work to obtain potential energy. Then you differentiate potential energy to obtain incremental work? But incremental work is given by the product of current force times [the parallel component of] incremental displacement. The force comes right out of Newton's law of gravitation.

[But perhaps I am not understanding your approach yet. Somewhere you decide to integrate potential energy -- which makes no sense. Are you, perhaps, thinking that incremental work is equal to potential energy instead of the differential thereof?]

If you do the differentiation correctly, you will get $dW=F\cdot dr$ where F is the same result that Newton would have given. Lots of work for an answer you already knew.

But all of that is irrelevant. You already know the potential energy. Forces add. The integrals of forces still add. So potential energies add. That's the principle of superposition. The potential energy of an object subject to two conservative fields is the sum of the potential energies from each field separately.

 

[Yalanhar]

Huh? What?

Someone else integrated work to obtain potential energy. Then you differentiate potential energy to obtain incremental work? But incremental work is given by the product of current force times [the parallel component of] incremental displacement. The force comes right out of Newton's law of gravitation.

[But perhaps I am not understanding your approach yet. Somewhere you decide to integrate potential energy -- which makes no sense. Are you, perhaps, thinking that incremental work is equal to potential energy instead of the differential thereof?]

If you do the differentiation correctly, you will get $dW=F\cdot dr$ where F is the same result that Newton would have given. Lots of work for an answer you already knew.

But all of that is irrelevant. You already know the potential energy. Forces add. The integrals of forces still add. So potential energies add. That's the principle of superposition. The potential energy of an object subject to two conservative fields is the sum of the potential energies from each field separately.

Yeah, I was trying to integrate the resultant gravitational force so I would get the potential energy already, instead of summing it. But I was not trying to integrate the potential energy, at least it wasn't my intention.

 

[PeroK]

Yeah, I was trying to integrate the resultant gravitational force so I would get the potential energy already, instead of summing it. But I was not trying to integrate the potential energy, at least it wasn't my intention.

I must admit I'm very confused by what you're doing. First, this variable $x$. Isn't it clear what value this has.

Second, the second part of the journey from $x$ to planet B. What is the relevance of this calculation?

 

[Yalanhar]

I must admit I'm very confused by what you're doing. First, this variable $x$. Isn't it clear what value this has.

Second, the second part of the journey from $x$ to planet B. What is the relevance of this calculation?

x is not a variable, is the point where $|F_a |=|F_b |$ So $x = \frac{R+D}{2}$
Yeah, I see now that there's no purpose on that

 

[PeroK]

x is not a variable, is the point where $|F_a |=|F_b |$ So $x = \frac{R+D}{2}$
Yeah, I see now that there's no purpose on that

Where are you measuring $x$ from?

 

[TSny]

The answer is $v_o = \sqrt \frac{GM(D-R)}{R(3R+D)}$

It's not clear to me whether D is the center-to-center distance of the planets or the surface-to-surface distance. But, either way, if you take the limit of D going to infinity, shouldn't the answer for $v_0$ just reduce to the escape velocity from A? But the expression given for the answer does not have this limit.

 

[PeroK]

Homework Statement: Two planets A and B (mass m for both) are separated by a distance of D. A object is thrown from the surface of planet A towards planet B. Radius of planet A and B are R and 2R. Find the minimum velocity to throw the object from planet A to B.
Homework Equations: $dW=\frac {GMm}{r^2}dr$
$W=\Delta U_p$
$K_o +U_o = K+U$

So I integrated the work done on the object by both planets. Work1 is until x, and Work2 is from x to d. Where x is the point where both gravitational forces are equal.

$W_1=\int_0^x \frac{GMm}{r^2}dr - \int_0^x \frac{GMm}{(3R+D-r)^2}dr$
$W_2=\int_x^D \frac{GMm}{(3R+D-r)^2}dr - \int_x^D \frac{GMm}{r^2}dr$

$W_t=W_1+W_2$

$W_t = \Delta U_p$

Chosing zero at A's surface

$W_t=U$

And then I have to conserve the energy and calculate $v_o$. However, the algebraic is a pain and I still have that $x=\frac{3R+D}{2}$.

Any help? The answer is $v_o = \sqrt \frac{GM(D-R)}{R(3R+D)}$

I agree with @TSny. For very distant planets, as D becomes very large, planet B has no significant influence in terms of escaping planet A.

Also, is the radius of planet B relevant? The expression you quote suggests $2R$ enters the equation at some point. Why should it?

 

[jbriggs444]

Also, is the radius of planet B relevant? The expression you quote suggests $2R$ enters the equation at some point. Why should it?

If the distance is measured surface to surface then the 2R would be expected to enter in. The "3R+D" in the denominator seems to suggest this, though I have not looked closely.

 

[Delta2]

It's not clear to me whether D is the center-to-center distance of the planets or the surface-to-surface distance. But, either way, if you take the limit of D going to infinity, shouldn't the answer for $v_0$ just reduce to the escape velocity from A? But the expression given for the answer does not have this limit.

The limit of the answer as $D\to\infty$ is just off by a factor of $\sqrt 2$ w.r.t the escape velocity from planet A. Probably it is a typo.

 

[Yalanhar]

Where are you measuring $x$ from?

From A

It's not clear to me whether D is the center-to-center distance of the planets or the surface-to-surface distance. But, either way, if you take the limit of D going to infinity, shouldn't the answer for v0v0v_0 just reduce to the escape velocity from A? But the expression given for the answer does not have this limit

Its the distance between the surfaces. But it doesn't say anything about D >> R.

I agree with @TSny. For very distant planets, as D becomes very large, planet B has no significant influence in terms of escaping planet A.

Also, is the radius of planet B relevant? The expression you quote suggests $2R$ enters the equation at some point. Why should it?

I think its relevant because D is not that greater than R.

 

[PeroK]

I think its relevant because D is not that greater than R.

Nevertheless the answer you are aiming for is wrong.

If D is small, then as mentioned above you have a very different problem.

I'm not sure what else to suggest.

 

[TSny]

The limit of the answer as $D\to\infty$ is just off by a factor of $\sqrt 2$ w.r.t the escape velocity from planet A. Probably it is a typo.

When I work out the answer I get a result that is a lot different than the answer given, not just a factor of $\sqrt 2$. Hope I'm not overlooking a mistake in my calculation.

 

[PeroK]

When I work out the answer I get a result that is a lot different than the answer given, not just a factor of $\sqrt 2$. Hope I'm not overlooking a mistake in my calculation.

Yes, and as I mentioned above there is no reason for terms like 3R to enter the expression, unless the radius of planet B, which is irrelevant, is used in some way.

 

[Highway_Dylan]

I think that the problem is intended to be solved with the assumption that $R$ isn't a lot smaller than $D$, but the motion of the planets is negligible during the time of flight of the projectile (say the planets are fixed somehow).

As was discussed above, the total potential energy is the sum of potential energies resulting from the interactions of the body with the planets (the potential resulting from the mutual interaction of the planets is constant by the above assumption and hence can be disregarded). You could start by writing the expression for the dependence of the potential energy on the position of the projectile measured from, say, the center of planet A. It may be helpful to plot the resulting expression. Looking at the plot, I believe that you could get an idea of how much kinetic energy you need.

All that being said, I don't believe that the given final solution is correct. It seems to be arriven at by inconsistent neglection of, or more probably, simply by forgetting, certain terms.

 

[TSny]

Yes, and as I mentioned above there is no reason for terms like 3R to enter the expression, unless the radius of planet B, which is irrelevant, is used in some way.

I do get the factors of R and (D+3R) in the denominator. The (D+3R) factor comes from the "midpoint" being (D+3R)/2 from the center of each sphere. But I also get an additional factor in the denominator and I get a different numerator. I keep looking for an arithmetical error in my calculation that would allow my extra factor in the denominator to cancel something in the numerator to end up with something close to the given answer. But no luck.

 

[Highway_Dylan]

I do get the factors of R and (D+3R) in the denominator. The (D+3R) factor comes from the "midpoint" being (D+3R)/2 from the center of each sphere. But I also get an additional factor in the denominator and I get a different numerator. I keep looking for an arithmetical error in my calculation that would allow my extra factor in the denominator to cancel something in the numerator to end up with something close to the given answer. But no luck.

The answer I got is:

$v_0^2=2GM\frac{(D+R)^2}{R(D+2R)(D+3R)}$.

Liable to my own algebraic error of course. I hope that posting this is allowed, since OP showed that he has the intended final answer available.

 

[Yalanhar]

Nevertheless the answer you are aiming for is wrong.

If D is small, then as mentioned above you have a very different problem.

I'm not sure what else to suggest.

I'll try what jbriggs444 suggested about adding energies

 

[TSny]

The answer I got is:

$v_0^2=2GM\frac{(D+R)^2}{R(D+2R)(D+3R)}$.

Liable to my own algebraic error of course. I hope that posting this is allowed, since OP showed that he has the intended final answer available.

Yes, that's the answer I get also, assuming that D is the distance between surfaces.

 

[Yalanhar]

The answer I got is:

$v_0^2=2GM\frac{(D+R)^2}{R(D+2R)(D+3R)}$.

Liable to my own algebraic error of course. I hope that posting this is allowed, since OP showed that he has the intended final answer available.

May I ask how?

 

[Highway_Dylan]

May I ask how?

Try what I (and also others) wrote in post #28. You can get the answer by using conservation of energy.

 

[PeroK]

The answer I got is:

$v_0^2=2GM \frac{(D+R)^2}{R(D+2R)(D+3R)}$

Latex sorted. We're all in agreement then.

@Yalanhar

In any case, if we are taking D to be the distance between the surfaces (which is really odd in a gravitational problem), then I would let $d = D +3R$ be the distance between the centers. Then work with $d$, as it is simpler and more standard, only converting to D for the final answer.