How to Calculate the Speed and Time of a Falling Object from Altitude?

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SUMMARY

The discussion focuses on calculating the speed and time of a falling object released from an altitude of 500 km above Earth's surface. The speed at a distance r from the Earth's center is defined by the equation v = sqrt(2GME (1/r - 1/(RE + h)). To find the time of fall, the integral (Delta) t = - (integral from i to f) dr/v must be performed numerically. The potential energy change is equated to the kinetic energy to derive the speed, confirming that M should be denoted as ME for clarity.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of the variables involved: G (gravitational constant), ME (mass of Earth), RE (radius of Earth)
  • Ability to perform numerical integration for solving differential equations
NEXT STEPS
  • Study the derivation of gravitational potential energy and its implications in physics
  • Learn numerical integration techniques, such as the trapezoidal rule or Simpson's rule
  • Explore the application of conservation of energy in gravitational systems
  • Investigate the effects of altitude on gravitational acceleration and its calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and gravitation, as well as educators seeking to explain the principles of motion under gravity.

prabhjyot
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need help please urgent

Homework Statement


An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth’s center, where RE < r < RE + h, is given by
v = sqrt(2GME (1/r -1/ (RE + h) )

(b) Assume the release altitude is 500 km. perform the integral:
(Delta) t = (integral from i to f) dt = - (integral from i to f) dr/v

to find the time of fall as the object moves from the release point to the Earth’s surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = -dr / dt. Perform the integral numerically.



The Attempt at a Solution



The Potential Energy at a disance h above Earth's surface (and therefore h+Re above Earth's center) = -GMm/(Re+h).

Potential at a distance r from center = -GMm/r
Change in PE = (-GMm/(Re+h)) - -(GMm/r) = GMm/r - GMm/(Re+h)

change in PE = Gain in KE = 1/2mv2

Solving for v

getting confused.../
 
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prabhjyot said:

Homework Statement


An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth’s center, where RE < r < RE + h, is given by
v = sqrt(2GME (1/r -1/ (RE + h) )

I presume here, ME is mass of the earth

(b) Assume the release altitude is 500 km. perform the integral:
(Delta) t = (integral from i to f) dt = - (integral from i to f) dr/v

to find the time of fall as the object moves from the release point to the Earth’s surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = -dr / dt. Perform the integral numerically.



The Attempt at a Solution



The Potential Energy at a disance h above Earth's surface (and therefore h+Re above Earth's center) = -GMm/(Re+h).

Potential at a distance r from center = -GMm/r
Change in PE = (-GMm/(Re+h)) - -(GMm/r) = GMm/r - GMm/(Re+h)

change in PE = Gain in KE = 1/2mv2

Solving for v

getting confused.../

Your method is correct. Where are you having problems solving for v? Set the KE equal to the PE (note that M shuld be ME in the notation in the question) and you more or less have the answer!

For the second bit, plug v into the integral.
 

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