Potential energy of a ball falling

The resulting integral cannot be done in terms of elementary functions, but you can evaluate it numerically using a computer or calculator.In summary, the speed of an object released from rest at an altitude h above the Earth's surface can be calculated using the formula v = sqrt(2GME (1/r -1/ (RE + h) ). To find the time of fall from the release point to the Earth's surface, the integral (Delta)t = (integral from i to f) dt = - (integral from i to f) dr/v can be used, where v is the calculated speed and ME is the mass of the Earth. However, this integral cannot be solved using elementary functions and must be evaluated numerically.
  • #1
prabhjyot
10
0
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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  • #2
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.
 

1. What is potential energy?

Potential energy is the energy possessed by an object due to its position or state. It is stored energy that has the potential to be converted into other forms of energy, such as kinetic energy.

2. How is potential energy of a ball falling calculated?

The potential energy of a ball falling is calculated using the formula PE = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height of the ball above the ground.

3. What happens to the potential energy of a ball as it falls?

As a ball falls, its potential energy decreases and is converted into kinetic energy. This is because the ball is moving closer to the ground, and therefore, its height (h) decreases in the potential energy formula.

4. Can potential energy of a ball falling be negative?

Yes, potential energy can be negative. This occurs when the ball falls below its initial height, resulting in a negative value for h in the potential energy formula. Negative potential energy indicates that the object is in motion and has less potential energy than its initial state.

5. Does the mass of the ball affect its potential energy when falling?

Yes, the mass of the ball does affect its potential energy when falling. The greater the mass, the greater the potential energy. This is because the potential energy formula includes the mass of the object (m), which directly influences the overall potential energy value.

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