Graphing a Trajectory with Variable Gravity

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SUMMARY

This discussion focuses on graphing the trajectory of an object launched from Earth's surface while accounting for the diminishing gravitational force as altitude increases. The relevant equations include the gravitational acceleration formula g(r / s)² and the position function x(t) = (1/2)*a*t² + v0*t + x0, with a specific example using escape velocity of 11184 m/s. The analysis progresses through differential equations to derive velocity and position over time, emphasizing that an object at escape velocity will not return to Earth due to reduced gravitational pull.

PREREQUISITES
  • Understanding of Newtonian physics and gravitational forces
  • Familiarity with calculus, particularly integration and differential equations
  • Knowledge of kinematic equations for motion
  • Basic understanding of graphing functions and trajectories
NEXT STEPS
  • Study the derivation of gravitational force variations with altitude
  • Learn about numerical methods for solving differential equations
  • Explore graphing software tools like Desmos or MATLAB for trajectory visualization
  • Investigate the implications of escape velocity in astrophysics
USEFUL FOR

Students in physics or engineering, educators teaching kinematics, and anyone interested in the mathematical modeling of motion under variable gravitational forces.

peterk
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Homework Statement


I would like to graph the position, velocity, and acceleration curves (over time) of an object that is launched straight up from the surface of the Earth, but I need to take into account the fact that the Earth's gravitational pull weakens as the object gets higher.

The position-time graph of an object launched at escape velocity should only go up and never down.

Homework Equations


g(r / s)2 = the acceleration of gravity at height s above the center of the Earth (where s > r ).
x(t)=(1/2)*a*t^2+v0*t+x0

The Attempt at a Solution


x(t)=(1/2)*-9.8*t^2+11184*t+0 (assuming that escape velocity is 11184). This would work except there is less gravity as the object moves farther away from the Earth, so it should never really return to the Earth's surface.
 
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From

[tex]a_y = ky^{-2}[/tex]

we progress to

[tex]\frac{d\ v_y}{dt} = ...[/tex]

and then to

[tex]\frac{d\ v_y}{dy}\ \frac{d\ y}{dt} = ...[/tex]

which gives

[tex]v_y\ \frac{d\ v_y}{dy} = ...[/tex]

integrating this gives

[tex]\int {v_y\ dv_y} = k\int {y^{-2}dy}[/tex]

...
 

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