Mass shot into orbit; find max distance from Earth

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Homework Help Overview

The problem involves a projectile of mass m being fired from the Earth's surface at a specific angle, with the goal of calculating the maximum height it reaches in relation to the Earth's radius. The scenario assumes no air resistance or Earth's rotation, and hints at using conservation laws related to the projectile's orbit.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the implications of the projectile's angle and initial speed on its trajectory and potential energy. There are considerations about comparing initial kinetic energy to that of a mass in orbit, and the need to understand the relationship between energy and angular momentum at different points in the projectile's path.

Discussion Status

Participants are exploring various aspects of energy conservation and the conditions for achieving maximum height. Some have suggested comparing energy and angular momentum at launch and apogee, while others express uncertainty about how to represent energy without knowing the mass of the projectile. The discussion is ongoing, with multiple interpretations being considered.

Contextual Notes

There is a noted lack of information regarding the mass of the projectile, which some participants mention as a constraint in their calculations. The problem also involves assumptions about the nature of the projectile's path and energy states at different points in its trajectory.

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Homework Statement


A projectile of mass m is fired from the surface of the Earth at an angle α from the vertical, where α = 0.46 radians. The initial speed v0 is equal to √{GM/Rearth}. Calculate the maximum height that the projectile will reach. Neglect air resistance and the Earth's rotation. Express the result as the ratio to the radius of the Earth. [Hint: Apply the conservation laws to the orbit of the projectile.]


Homework Equations



E=\frac{1}{2}m\dot{r}2+Ueff(r)

Ueff(r)=\frac{L^2}{2mr^2}+U(r)

The Attempt at a Solution



Intuitively speaking the mass will go into orbit and because we're not necessarily near the surface of the Earth the potential is not mgh. Also, the orbit must be bound or the answer would be infinity. If the mass is in orbit I could compare the effective potential of two points in the orbit but I don't see how to make a useful comparison with the information I am provided. What would be the variable of comparison?

Should I be comparing the initial kinetic energy to that of a mass in orbit? I can't seem to figure out how to cancel out v02 if that is the case:

\frac{1}{2}mv02=\frac{1}{2}m\dot{r}2+Ueff(r)

Any suggestions? Thank you for your help.
 
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If the object is fired from the surface at some angle other than zero, then eventually the object will impact back into the earth, unless it reaches or exceeds escape velocity.

Getting back to the problem statement, you're given the initial energy (potential + kinetic) and the angle. You need to determine when the potential energy is at a maximum and kinetic energy is at a minimum, which will correspond to the maximum height. You may need to determine the parameters of the elliptical path.
 
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I was given the speed. I don't have the object's mass. I don't know how to represent its energy in terms of v0 besides 1/2mv^2 which won't work here.

The potential is at a maximum at the apogee. How I know where the apogee is I am not sure.

E=\frac{1}{2}mv2-\frac{GMm}{r}
 
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Try comparing energy (KE+PE) and angular momentum about Earth's centre between launch and apogee. You won't need to know the mass - that will cancel out.
 

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