Gravitation and its effect on elliptical orbits

In summary, the conversation discusses the calculation of a comet's speed when at two different distances from the sun, using the formula for velocity in a circular orbit and the concept of equal areas swept out in equal times. The resulting equations and steps are shown, with some trial and error involved in reaching the correct answer.
  • #1
marekkpie
6
0

Homework Statement



Comets travel around the sun in elliptical orbits with large eccentricities. Suppose the comet has an initial speed of 1.17*10^4 m/s when at a distance of 4.9*10^11 m from the center of the sun, what is its speed when at a distance of 5.9*10^10 m? Give your in m/s in scientific notation to three significant digits. (Note: Use appendix F for the necessary data.)

Homework Equations



Don't know 'em. All I know is finding velocity in a circular orbit:

v = sqrt((Gravitational constant * mass of the object being orbited around) / radius from the center of the orbited object)

The Attempt at a Solution



v = sqrt((6.67E-11 * 1.99E30) / 5.9E10) = 4.74E4

The correct answer is: 6.40E4...so since I'm in the same order of magnitude I assume I'm close.
 
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  • #2
Orbits sweep out equal areas in equal times
Consider the area of a triangle swept by the comet in 1 sec (eg some time short enough that you can make the path a straight line) when it's far and close to the sun.

If the area is the same and you know the height you cna work out how long the base is and so the speed.
 
  • #3
O.K...so since the times and areas are equal, we can safely set both sides to equal the area of a triangle? i.e.

.5 * R1 * V1 = .5 * R2 * V2
thus,
V2 = V1 * R1 / R2

But then I get 9.72E4.
 
  • #4
mgb_phys said:
Orbits sweep out equal areas in equal times
Consider the area of a triangle swept by the comet in 1 sec (eg some time short enough that you can make the path a straight line) when it's far and close to the sun.

If the area is the same and you know the height you cna work out how long the base is and so the speed.

There was no mention of
perihelion or aphelion.

Conservation of energy is what is expected

David
 
Last edited:
  • #5
Alright, then here is what I got from davieddy's help:

K + U = K + U

MC = mass of the comet

MS = mass of the sun

G = gravitational constant

So...here is my equation, followed by the steps to make it solve for V2:

.5 * MC * V1 ^ 2 - G * MS * MC / D1 = .5 * MC * V2 ^ 2 - G * MS * MC / D2
<=>
.5 * V1 ^ 2 - G * MS / D1 = .5 * V2 ^ 2 - G * MS * D2
<=>
...bunch of algebra...
<=>
V2 = (V1 ^ 2 + .5 * G * MS (D2 ^ -1 - D1 ^ -1)) ^ .5

which leaves me with 3.36E4. Still in the right ballpark, but not quite there.
 
  • #6
marekkpie said:
V2 = (V1 ^ 2 + .5 * G * MS (D2 ^ -1 - D1 ^ -1)) ^ .5

which leaves me with 3.36E4. Still in the right ballpark, but not quite there.

Try
V2 = (V1 ^ 2 + 2 * G * MS (D2 ^ -1 - D1 ^ -1)) ^ .5
 
  • #7
davieddy said:
Try
V2 = (V1 ^ 2 + 2 * G * MS (D2 ^ -1 - D1 ^ -1)) ^ .5

D'oh! Tyvm.
 

FAQ: Gravitation and its effect on elliptical orbits

1. How does gravitation affect elliptical orbits?

Gravitation is the force that governs the motion of objects in space. In the case of elliptical orbits, the gravitational force from a central body, such as a planet or star, is what causes an object to follow a curved path. The strength of the gravitational force determines the shape and size of the elliptical orbit.

2. Why do planets and other celestial bodies have elliptical orbits?

According to Newton's law of gravitation, the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In the case of planets orbiting a star, the gravitational force decreases as the distance between them increases. This results in an elliptical orbit as the planet moves closer to and further away from the star.

3. Can gravity affect the speed of an object in an elliptical orbit?

Yes, gravity can affect the speed of an object in an elliptical orbit. As an object moves closer to the central body, the gravitational force increases and speeds up the object. As the object moves further away, the gravitational force decreases and slows down the object. This results in a varying orbital speed throughout the elliptical orbit.

4. How do scientists calculate the gravitational force in an elliptical orbit?

To calculate the gravitational force in an elliptical orbit, scientists use the equation F = (G * m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. This equation allows scientists to determine the strength of the gravitational force at any point in the orbit.

5. Can objects in elliptical orbits eventually escape the gravitational pull of a central body?

Yes, objects in elliptical orbits can eventually escape the gravitational pull of a central body. This can happen if the object's speed is great enough to overcome the gravitational force pulling it towards the central body. This is known as escape velocity and varies depending on the mass of the central body and the distance of the object from it.

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