# Newton's law of universal gravitation

1. Sep 15, 2013

### Permanence

1. The problem statement, all variables and given/known data
A star of mass 5 × 10e30 kg is located at ‹ 7 × 10e12, 3 × 10e12, 0 › m. A planet of mass 4 × 10e24 kg is located at ‹ 5 × 10e12, 5 × 10e12, 0 › m and is moving with a velocity of ‹ 0.6 × 10e4, 1.4 × 10e4, 0 › m/s.

A. During a time interval of 1x10e6 seconds, what is the change in the planet's velocity?
B. During this time interval of 1x10e6 seconds, what is the change in the planet's position?

2. Relevant equations
- Newton's Law of Gravity
- Unit Vector = (Vector) / (Magnitude of Vector)
- Momentum Final = Momentum Initial + Force Net * Δt = mass*Velocity initial + Force net Δt

3. The attempt at a solution

I began by listing the variables I was given. I then subtracted the position of the star from the position of the planet.
r = r(planet) - r(star) = ( -2e12, 2e12, 0)

I then found the magnitude of the vector r. |r| = 2.828e12

I then found the force of gravity between the planet and star using Newton's Law of Gravity.
F(grav) = 1.67e20

I then found the unit vector of r with:
Unit Vector = Vector / Magnitude of Vector
r(hat) = (-0.707, 0.707, 0)

I then multiplied my unit vector by F(grav) to obtain the components. I got the values:
(-1.18e20, 1.18e20, 0)

I then listed the formula:
Momentum final = velocity initial * mass of planet + Fnet * Δt
I obtained the value (2.4e48, 5.61e28, 0)

I then used the following formula: v(final) = momentum final / mass
I got the following value: (6e23, 14025, 0)

I then subtracted my initial velocity from my final velocity and obtained:
Δv = (6e29, 1.0009e13)

To approximate my new position I used: Position final = position initial + v(avg)Δt ; where v(avg) = ( 3e29, 2.505e12, 0)
Position final = (3e35, 2.505e18, 0)
I then subtracted my final position from my initial position to obtain:
Δposition = (2.999e35, 2.504e18,0)

I did not receive any points for my answer. I have attempted this problem a few times with no success. I have also attempted sample problems (which have the final solution) but my answer does is not remotely close. I'd really appreciate some advice or someone to review my work.

With Regards,
Permanence

Last edited by a moderator: May 6, 2017
2. Sep 15, 2013

### lendav_rott

What is the product of gravity and the unit vector supposed to be about? Another thing is, atleast seems odd to me that you went through with it: The planet is in constant motion hence the direction of this imaginary vector of velocity is constantly changing, I don't think you can just perform simple addition here.

Final velocity seems kind of a utopia to me - your planet starts at a range of 10^4 m/s in both x and y direction and now suddenly after a bit of time it has accelerated to 10^29 and 10^13 x and y - is that how the Earth will end up? :O

By the way is e supposed to mean 10^ or just the power sign, I am confused.

Last edited: Sep 15, 2013
3. Sep 15, 2013

### Permanence

I don't think there is anything wrong with me finding the component vector of gravity by multiplying by the unit vector. Could someone else possibly weigh in on this?

@lendav_rott, I appreciate you taking your time to respond. I assume that since you responded you understand how to solve the problem. If that is the case, could you perhaps give me a starting point?

4. Sep 15, 2013

### lendav_rott

This assignment is built on a phenomenon called centripetal force. Read up on it :)

If you tie an object to a rope and swing it around you like it was a planet and you were the star the object's resulting velocity only changes if you start going round slower or faster. What does that mean in your planet-star situation?

..the planet's resulting velocity will Not change, but the x y components will, which you are kindly asked to find at a given time stamp.
Whenever you solve something, think about the result - the speed of light is around the range 10^8 m/s, your planet has long ago broken the speed of light which is, I think you will agree, impossible.

Do you know how to calculate rotation of a given point on a x,y plane that circles another given point?

The assignment does not say that the planet is circling the star but I am assuming it does and since there is no further info on the orbit I asssssume it is circular.

Last edited: Sep 15, 2013
5. Sep 15, 2013

### Permanence

Lol you make a good point. I didn't realize it broke the speed of light. We have not formally gone over the stuff you mentioned. I solved the problem almost exactly how our book solved a similar problem.

When I click the show solution of the practice problem it says to do the following:
"1) Calculate the gravitational force.
2) Update the momentum. The change in the momentum is equal to the impulse; divide by the planet's mass to find the change in velocity.
3) Update the position. The change in the position is approximately the final velocity times the time interval."

6. Sep 15, 2013

### lendav_rott

I would still like to see your reasoning for that Force*unit vector - N*m is a unit for Work.

To whatever end you have to find out what the angle is between the imaginary trajectory of the planet's starting point and position after t = 10^6 seconds.
The motion is circular therefore the book's method will not be enough.

Position final = (3e35, 2.505e18, 0)
this is what you arrived at after all - the distance between the star and the planet is 10^12 and now your planet has wandered off into space somewhere and travelling at unimaginable speeds to boot.

When you calculate the final velocity of the planet you will get an answer that is 41.75 m/s faster than the initial velocity, but it's negligible really, because the initial velocity is roughly 1.52*10^4 m/s, what's a few m/s here n there really. Since the book is so ineloquent in that regard, we might aswell consider this as uniform circular movement in which case:
F = m|r|ω2 where ω is the angular velocity - rad/s, m is the planet's mass and |r| the distance from the star.
also:
a=V²/|r|
and of course, F = ma
ω*t will be the angle you are looking for - mind you the result will be in radians - call it θ

Calculating the positions will mean you are rotating the r vector around the star. The planet is moving around the star clock-wise, sketch it on paper and you will see.

to find the rotated (clockwise) vector's coordinates
r'x = rx*cosθ + ry*sinθ
r'y = rx*sin(-θ) + ry*cosθ

You will add these coordinates to the star's position and will arrive at where the planet will be after 10^6 seconds.
Similarly you can find what the velocity components are.

It seems kind of unbelievable to me they would have you tackle such assignments without going over the material concerning a Central idea to the whole shebang here. If you're stuck show where you are at and we'll figure it out.

Cheers

Last edited: Sep 16, 2013
7. Sep 15, 2013

### Permanence

Thank you for responding. I will attempt to rework the problem with the information you have provided me and will reply within an hour.

The reason I did that was because in our book they worked through a similar problem by doing it:
http://gyazo.com/aeec6ef78ced05e7734e001953e4792d.png [Broken]

Last edited by a moderator: May 6, 2017
8. Sep 15, 2013

### nasu

Some of your numerical calculations are way off.
The initial momentum of the planet is of the order 10^28 Ns and the impulse of the force is of the order 10^26 Ns. There is no way to get 10^48 in the final momentum.

I mean in the original post.
I suppose is just and error of calculation.

9. Sep 16, 2013

### lendav_rott

When you perform an operation ask yourself what you think it is necessary for. Calculating the unit vector of r has a use and it helps you disect the Force vector into x, y components, because r and F always are coinciding with one another, but why do you think it is F*(-r) in the book?

According to the book, were you to calculate the final velocity x component:
Vxfinal = Vxinitial + Fx* t/mplanet
That is:
Vxfinal = 6000 + 1.18*10^18/(4*10^24) = ~0.6*10^4 because the change is so negligibly small, it's 2.95*10^-7 difference and so similarly you will calculate the y component
Vyfinal = 14000 -1.18*10^18/(4*10^24) = same deal, the difference is ever so small.

When you use the method I showed you, you will have to reach a similar result depending on your rounding habits.

And it makes sense that the velocities change only slightly, the distance from the star is about 2.82*10^12 m, Earth is 1.5*10^12 m from the Sun, 1mil seconds is about 11.5~ days, the planet is moving slower than the Earth further away from its star so there really won't be much change in the position either. If the position of the planet were to change radically, it means the velocity components would also have to change a lot and vice versa, because the resulting velocity vector and the Force vector always have a right angle between them so to make up for the ever changing direction of velocity, the magnitudes of the x, y components will change accordingly.

What is this un-usable result at t=10^9s about? I really don't see, the planet will have travelled for 11574 days roughly and has probably circled the star for quite a few times by then, but that is no reason for it to produce unusable results.

You have to bombard your teacher with questions, after all, it's what they are for. :D

Last edited: Sep 16, 2013
10. Sep 16, 2013

### Staff: Mentor

I thought I'd make a few observations on the orbit of the planet described in the problem, for no particular reason

Given the mass of the star and the initial position and velocity of the planet we can quickly work out a few details about the orbit. Suppose that $\vec{r_o}$ and $\vec{v_o}$ are the initial position and velocity vectors respectively, and $r_o$ and $v_o$ are their magnitudes. In this case we have:

$\vec{r_o} = (-2.00 \hat{i} + 2.00 \hat{j} + 0.00 \hat{k}) 10^{12} m = (-13.4 \hat{i} + 13.4 \hat{j} + 0.00 \hat{k})\; AU$
$\vec{v_o} = (0.6 \hat{i} + 1.4 \hat{j} + 0.0 \hat{k}) ms^{-1}$

$r_o = 2.83 x 10^{12}m\;\approx 19 AU$
$v_o = 1.52 x 10^4 m/s$

A few relevant parameters:

$\mu = G*M_{star} = 3.34x 10^{20} m^3s^{-2}~~~~~$Gravitational Parameter for the system
$\xi = \frac{v_o^2}{2} - \frac{\mu}{r_o} = -2.00 x 10^6 J/kg~~~~$ Negative Specific Energy implies the orbit is closed (The planet is bound)
$a = \frac{-\mu}{2\xi} = 8.40 x 10^{12} m = 562 AU~~~~$ Wow! That's a large orbit; very elliptical given starting radius
$\vec{h} = \vec{r_o}\times \vec{v_o} = -4.00 x 10^{16}\; \hat{k}\; m^2s^{-1}~~~~$ Specific angular momentum points along the negative z-axis, so the orbit is clockwise.
$\vec{e} = \frac{\vec{v_o} \times \vec{h}}{\mu} - \frac{\vec{r_o}}{r_o} = -0.971 \hat{i} + 0.012 \hat{j} + 0.00 \hat{k}~~~~~$ Eccentricity vector -- points in direction of periapsis
$e = |\vec{e}| = 0.971~~~~~$ The eccentricity is high, orbit is getting close to parabolic
$TP = \frac{2\pi}{\sqrt{\mu}}a^{3/2} = 2.645 x 10^{11}s \approx 8400\; yrs~~~~$ Orbital period
$\nu_o = acos\left(\frac{\vec{r_o}\cdot\vec{e}}{r_o e}\right) = 44.3°~~~$ True Anomaly (orbital angle since perihelion)

So as we can see it's a pretty large obit: a rather stretched-out ellipse and the given initial position is fairly close to perihelion. The orbital period is on the order of 8400 years.

Enjoy.

Last edited: Sep 16, 2013
11. Sep 16, 2013

### lendav_rott

Wow, nice one, although I doubt that one is required to know all of this stuff for this assignment. Interesting nonetheless ^^