Launching a Satellite: Finding Velocity and Max Distance

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Homework Help Overview

The problem involves a satellite launched from a distance of 2R from the center of the Earth, at an angle of 60° to the vertical, and its subsequent trajectory leading to a collision with the south pole. The focus is on determining the launch velocity and the maximum distance from the Earth's center based on principles of energy and angular momentum conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of energy conservation and angular momentum conservation to derive the launch velocity. There are questions regarding the correctness of the potential energy terms and the implications of angular momentum at the point of collision. Some participants also explore the relationship between the position and velocity vectors and the use of polar coordinates for trajectory analysis.

Discussion Status

There is an ongoing examination of the equations used, with some participants providing corrections and alternative perspectives on the setup. The discussion reflects a mix of interpretations regarding the trajectory and the role of mass in the equations. Guidance has been offered on considering additional equations beyond conservation laws.

Contextual Notes

Participants note potential misunderstandings regarding gravitational potential energy and the assumptions made about the angle of impact at the south pole. The discussion also highlights the challenge of deriving trajectory information without knowing the mass of the satellite.

AwesomeTrains
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Homework Statement


R is Earth radius.
A satellite is launched 2R from the center of the earth, vertically above the northpole, at an angle of 60° to the vertical.

The satellite crashes on the southpole. Find the launch velocity and the maximum distance to the center of the Earth from its trajectory.

Homework Equations


Energy conservation.
Angular momentum is also conserved.

The Attempt at a Solution


I tried using those two conservation laws to first get the launch velocity.
Energy conservation:
[itex]\frac{mv^{2}}{2}+\frac{GMm}{2R}=\frac{mu^{2}}{2}+\frac{GMm}{-R}[/itex]
Angular momentum conservation:
[itex]v \cdot 2R \cdot sin(Pi/3) = u \cdot -R[/itex]

Solving first equation for u and plugging into the second gives me [itex]9684.37 m/s[/itex]

Can I find the maximum distance by using the same equations?
 
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hold on, hold on. In your equation for energy conservation, on the left hand side, you've written the potential energy as positive. But gravitational potential energy can't be positive. Maybe this is a typing mistake? Also, in your equation for angular momentum conservation, you have written the final angular momentum as ##-Ru## but this would imply that the satellite has zero radial velocity when it collides with the Earth (which definitely can't be true). You got the left hand side correct, since you've used the correct angle there.
 
BruceW said:
hold on, hold on. In your equation for energy conservation, on the left hand side, you've written the potential energy as positive. But gravitational potential energy can't be positive. Maybe this is a typing mistake? Also, in your equation for angular momentum conservation, you have written the final angular momentum as ##-Ru## but this would imply that the satellite has zero radial velocity when it collides with the Earth (which definitely can't be true). You got the left hand side correct, since you've used the correct angle there.

I rewrote the energy conservation equation:
[itex]\frac{-MGm}{2R}+\frac{v^{2}}{2}=\frac{-MGm}{-R}+\frac{u^{2}}{2}[/itex]
What is the angle when it hits the southpole? Isn't it 90°?
 
uh, hehe. now the right hand side isn't correct. it is the magnitude of the displacement that enters into the equation for potential. At the south pole, the displacement will be ##-R\hat{z}## (where the z hat is unit vector). The magnitude of this is not ##-R##.

when it hits the south pole, why would the angle be 90° ? Maybe you are thinking of problems where the satellite 'just touches' the earth. But this question does not mention that.
 
Ah okay, now I got the energy right, I think.
[itex]\frac{mv^{2}}{2}-\frac{GMm}{2R} = \frac{mu^{2}}{2} - \frac{GMm}{R}[/itex]

But I have no idea on how to find the angle between the position vector and the velocity vector.
Is it possible to find the trajectory when you don't know the mass of the satellite? And should I use polar coordinates?
 
AwesomeTrains said:
Ah okay, now I got the energy right, I think.
[itex]\frac{mv^{2}}{2}-\frac{GMm}{2R} = \frac{mu^{2}}{2} - \frac{GMm}{R}[/itex]

But I have no idea on how to find the angle between the position vector and the velocity vector.
Is it possible to find the trajectory when you don't know the mass of the satellite? And should I use polar coordinates?

Look carefully at your equation. Do you notice any common factors?
 
Yea, m. But can I get the trajectory from the energy equation?
 
hmm. good point, I don't think you can do this problem with just the equations for conservation of energy and conservation of angular momentum alone. try using more equations!
 
But at least you know the mass of the satellite is immaterial.
 
  • #10
For the trajectory:

It may be a bit tedious, but you should be able start with the polar form of the ellipse equation for an orbit. The plot sits on X--Y axes with the focus at the origin. Place the "Earth" centered on the origin, too. The Earth's polar axis will be rotated through some angle θ to avoid forcing a rotation on the ellipse equation. (Suggestion: Define the Earth radius to be 1 unit so you don't have to carry R through all the manipulations).

You can find equivalent Cartesian coordinates for points on the ellipse readily enough, and so you can determine the slope at a strategic location...

Find the semi latus rectum and the eccentricity and everything else will fall into place.
 

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