Gravitational forces, speeds and masses

In summary, a comet moves around a star in a elliptical orbit described by 0.16 x^2 + y^2 = 4where x, y are in AU.The comet has an eccentricity of 0.917 and a semi-major and minor axis of 5 and 2 AU, respectively. The perihelion has a speed of 100 km/s and the aphelion has a speed of 5 km/s. The comet's mass is estimated to be an order of magnitude less than the star's mass.
  • #1
gelfand
40
3

Homework Statement



A comet moves around a stat in ##xy## plane along elliptical orbit, described by

$$
0.16 x^2 + y^2 = 4
$$

where ##x, y## are in ##AU##

1) Sketch the comet in the ##x,y## coordinate system denoting all orbit parameters

2) Find the semi major and minor axes of the orbit in ##AU##

3) Assuming the star is located at ##x < 0##, find the eccentricity of the orbit
and the ##x, y## coordinates of the star

4) If the orbital speed of this comet at the perihelion is ##100 ## km/s, find
its orbital speed at aphelion

5) Using the energy conservation approach find the approximate mass of the star
to an order of magnitude.

Homework Equations

Eccentricity ##e = \sqrt{1 - \frac{b^2}{a^2}}##f ## = \sqrt{a^2 - b^2}## , I'm not sure what this is other than the distance
between where the semi-major and minor axis meet and the location of the star.

The Attempt at a Solution

For part 1) I have drawn a sketch.

For part 2) I note that the given expression is $$
0.16 x^2 + y^2 = 4
$$

Plugging ##x = 0## into this gives ##y^2 = 4## so ##y = 2##. Then setting ##y = 0##
gives ##x^2 = \frac{4}{0.16}## so ##x = 5##.

This gives us our semi-major axis at ##5## and the semi-minor at ##2##, both in ##AU##.

Part 3) Using the formula for eccentricity as

$$
e = \sqrt{1 - \frac{4}{25}} = \frac{\sqrt{21}}{5} \approx 0.917
$$

The ##x, y## coordinates of the star are found using ##f = \sqrt{a^2 - b^2}## which
is

$$
f = \sqrt{25 - 4} = \sqrt{21} \approx 4.583
$$

From this we note that the location of the star is ##(-4.583, 0)## to 3 decimal
places.

part 4)

I know that the perihelion has the fastest speed - but I'm not sure how to go
about finding the speed of the aphelion given this information.

Should I be setting up some kind of differential equation? Or is there an
appropriate formula?

part 5)

I'm not sure how to consider this either. If the energy is conserved then I have
the definition

$$
W + PE_0 + KE_0 =
PE_f + KE_f + \text{Energy(Lost)}
$$

Given that there's no work being done in the motion and there's no energy lost
due to friction we have

$$
PE_0 + KE_0 =
PE_f + KE_f
$$

I'm not sure what to do with this though, in terms of finding the mass.

There's no 'at rest' for this system is there, in the way that this expression
is often used.

The potential energy would be a result of the gravitational field, which is

$$
\text{Gravitational force } = \frac{-GMm}{x}
$$

Where ##x## is the distance between the orbiting body and the star, ##G## is the
gravitational constant, ##M## is the mass of the star (what we want) and ##m## is
the mass of the orbiting body (which we don't have either).Thanks
 
Physics news on Phys.org
  • #2
For part (4) consider angular momentum.
 
  • #3
gneill said:
For part (4) consider angular momentum.

cheers - so for part 4 if I consider the angular momentum ##\omega## then:

##v = \omega r##, and here we have ##r = a##, the semi-major axis and the speed is
##100##km/s.

So we have

$$
\frac{v}{r} = \omega
$$

Which gives us ##\omega = \frac{100}{5} = 20##.

But this doesn't take into consideration the position of the star?
I've not used any information about the mass, star position, just treated it
like it was a uniform circular motion (i think?)
 
  • #4
gelfand said:
But this doesn't take into consideration the position of the star?
I've not used any information about the mass, star position, just treated it
like it was a uniform circular motion (i think?)
Right. The motion is about the star and the velocity is only perpendicular to the radius at the ends of the major axis (The scalar version of ##\vec{h} = \vec{r} \times \vec{v}##, namely h = rv, needs r and v to be at right angles just as in circular motion). That means you need to use the perihelion and aphelion distances, so they want you to calculate those first (Don't worry, you'll need them for part 5 as well ).
 
  • #5
OK - so I have that ##f =\sqrt{21}##, and this is the distance from the star to
the 'origin' (where major / minor meet). So from this I can say that the
distance from the star to the orbiting body is

$$
dist = major - f
$$

Which here is

$$
dist = 5 - \sqrt{21} = 0.417242305
$$

So that I have the perihelion distance as ##\approx 0.4172##

The aphelion distance would be (I think)

$$
2 \times \text{semi major} - (\text{perihelion distance})
$$

Which here is

$$
10 - 0.4172 = 5 + \sqrt{21}
$$

So I have these two distances, and need to use the angular momentum to find the
mass of the star.

I'm still considering angular momentum to be ##\omega## which is defined as
##\omega = \frac{\theta}{T}## where ##\frac{1}{t} = f = ## frequency.

But from this - I'm not sure what the frequency really is?

Perhaps If I use

$$
v = \omega r
$$

Then I can sub ##r## for the perihelion distance and express this as

$$
100 = \omega \times (\text{perihelion distance}) = \omega \times (5 - \sqrt{21})
$$From this I have that

$$
\frac{100}{5 - \sqrt{21}} = \omega \approx 239.5643924
$$

So this is my angular momentum , which I may presume is a constant.

Then If I have this as ##\omega## I can use it to find speed as

$$
v = \omega \times r
$$

Then

$$
v = 239.5643924 \times (5 - \sqrt{21}) \approx 2285.64
$$

Ouch, this is way off, as the speed must be slower here.

I'm confused.
 
  • #6
gelfand said:
OK - so I have that ##f =\sqrt{21}##, and this is the distance from the star to
the 'origin' (where major / minor meet). So from this I can say that the
distance from the star to the orbiting body is

$$
dist = major - f
$$

Which here is

$$
dist = 5 - \sqrt{21} = 0.417242305
$$

So that I have the perihelion distance as ##\approx 0.4172##

The aphelion distance would be (I think)

$$
2 \times \text{semi major} - (\text{perihelion distance})
$$

Which here is

$$
10 - 0.4172 = 5 + \sqrt{21}
$$

So I have these two distances, and need to use the angular momentum to find the
mass of the star.
Okay, your two distances look good.
I'm still considering angular momentum to be ##\omega## which is defined as
##\omega = \frac{\theta}{T}## where ##\frac{1}{t} = f = ## frequency.
##\omega## is angular velocity. For a point mass, angular momentum is given by ##\vec{L} = m \vec{r} \times \vec{v}##.

Usually for orbital mechanics we consider "specific" quantities (per unit mass) since we often don't know or care about the mass of the orbiting object as it's usually insignificant with respect to the mass of the primary. So, dropping the "m" the specific angular momentum is given by

##\vec{h} = \vec{r} \times \vec{v}##

When ##\vec{r}## and ##\vec{v}## are perpendicular (at the ends of the major axis), you have the scalar relationship ##h = r v##.
 
  • #7
Thanks - are you just using ##\vec{h}## here to represent angular momentum for
this case? (I've not seen ##\vec{h}##, just making sure that's what this was for
here).

Then from this we have that

$$
h = rv
$$

So for this problem we have, for the perihelion position (where the object is closest)

$$
h = 100 \times (5 - \sqrt{21}) \approx 41.7424
$$

So then at the other end I have that the speed is

$$
41.7424 = v \times (5 + \sqrt{21})
$$

Then

$$
v = \frac{41.7424}{5 + \sqrt{21}} \approx 4.35607
$$Therefore the speed at the aphelion position would be ##v = 4.35607##.

For part 5 then I need to consider the mass of the object.

I'm not sure what equations I should use for this.
 
  • #8
gelfand said:
Thanks - are you just using ##\vec{h}## here to represent angular momentum for
this case? (I've not seen ##\vec{h}##, just making sure that's what this was for
here).
Yes. h is the commonly used variable for the specific angular momentum in orbital dynamics.
Then from this we have that

$$
h = rv
$$

So for this problem we have, for the perihelion position (where the object is closest)

$$
h = 100 \times (5 - \sqrt{21}) \approx 41.7424
$$

So then at the other end I have that the speed is

$$
41.7424 = v \times (5 + \sqrt{21})
$$

Then

$$
v = \frac{41.7424}{5 + \sqrt{21}} \approx 4.35607
$$Therefore the speed at the aphelion position would be ##v = 4.35607##.
Good. Be sure to tack on the appropriate units and round to an appropriate number of sig figs.
For part 5 then I need to consider the mass of the object.

I'm not sure what equations I should use for this.
The mass of the object will drop out along the way if you choose to include it in the workings. The problem gave you the hint of using an energy conservation approach. What energy conservation formula applies to orbits?
 
  • #9
So we have

$$
\frac{1}{2}mv_1^2 - \frac{GMm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GMm}{r_2}
$$
Here we want to solve for ##M##, and can see that every term has ##m## in it (nice!)
so can just drop that :)

Dropping ##m## and multiplying through by 2 gives

$$
v_1^2 - \frac{2GM}{r_1} = v_2^2 - \frac{2GM}{r_2}
$$

Then

$$
v_1^2 - v_2^2 =
\frac{2GM}{r_1}
- \frac{2GM}{r_2}
$$

So

$$
M = \frac{v_1^2 - v_2^2}{\frac{2G}{r_1} - \frac{2G}{r_2}}
$$

And from here it's a case of plugging in the known values.

I'm not too sure which ##r## is which here though.
I think that I would have

##r_1## as the perihelion distance and ##r_2## as the aphelion distance (from the star).
 
  • #10
gelfand said:
I'm not too sure which ##r## is which here though.
I think that I would have

##r_1## as the perihelion distance and ##r_2## as the aphelion distance (from the star).
Yup.
 
  • #11
gneill said:
Yup.
Thanks - so that's this question done then?
 
  • #12
gelfand said:
Thanks - so that's this question done then?
Looks like it to me. What final value did you obtain for the mass of the star?
 

Related to Gravitational forces, speeds and masses

What is the difference between gravitational force and gravitational acceleration?

Gravitational force is the attractive force between two objects with mass, while gravitational acceleration is the rate at which an object falls towards a larger mass. In other words, gravitational force is a force acting between two objects, while gravitational acceleration is an acceleration acting on a single object.

How does mass affect gravitational force?

Mass directly affects gravitational force, as the larger the mass of an object, the greater the gravitational force it exerts. This is because gravitational force is directly proportional to the masses of the two objects involved.

What is escape velocity and how is it related to gravitational force?

Escape velocity is the minimum velocity needed for an object to escape the gravitational pull of a larger mass. It is related to gravitational force because the greater the gravitational force of the larger mass, the higher the escape velocity needs to be.

How do different speeds affect gravitational force?

The speed of an object does not directly affect gravitational force, but it can affect the gravitational potential energy of the object. As an object gains speed, its gravitational potential energy decreases, which can in turn affect its gravitational force.

What is the equation for calculating gravitational force between two masses?

The equation for calculating gravitational force is F = G(m1m2)/r^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of the two objects.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
552
  • Introductory Physics Homework Help
Replies
3
Views
967
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
276
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
762
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top