How Accurate Is the Calculation of Potential Energy Change at High Altitudes?

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Homework Help Overview

The discussion revolves around calculating the change in potential energy for a mountaineer ascending Mount Everest, focusing on the implications of using a constant gravitational acceleration versus a variable one at high altitudes.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of potential energy using the formula ΔEp=mgΔh, while questioning the accuracy of using a constant value for gravitational acceleration at high altitudes.
  • Some participants suggest integrating the gravitational force over the height to obtain a more accurate potential energy change, referencing Newton's law of gravitation.
  • There are inquiries about the integration process and the mathematical steps involved in calculating the potential energy change accurately.

Discussion Status

The discussion is ongoing, with participants sharing their attempts at integrating the gravitational force and expressing uncertainty about the mathematical methods involved. There is no explicit consensus, but some guidance on the integration process has been provided.

Contextual Notes

Participants note the challenge of accurately calculating potential energy change at high altitudes due to the variation in gravitational acceleration, raising questions about the assumptions made in standard calculations.

studentxlol
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Homework Statement



Calculate the gain of potential energy of a mountaineer of mass 80kg who travels to the top of the mountain Everest 9km above sea level.


Homework Equations



ΔEp=mgΔh

The Attempt at a Solution



ΔEp=mgΔh=80x9.81x9x10^3=7.1x10^6J.

This is the correct answer but surely it's only an estimate since the value of g 9km above the Earth's surface is 9.83830. Δg=9.8380-9.81= 2.8x10^(-3)?

Is it possible to calculate the exact change is Ep as the mass moves through Δh, rather than using a fixed value of 9.81?
 
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studentxlol said:

Homework Statement



Calculate the gain of potential energy of a mountaineer of mass 80kg who travels to the top of the mountain Everest 9km above sea level.


Homework Equations



ΔEp=mgΔh

The Attempt at a Solution



ΔEp=mgΔh=80x9.81x9x10^3=7.1x10^6J.

This is the correct answer but surely it's only an estimate since the value of g 9km above the Earth's surface is 9.83830. Δg=9.8380-9.81= 2.8x10^(-3)?

Is it possible to calculate the exact change is Ep as the mass moves through Δh, rather than using a fixed value of 9.81?
Yes. One would need to go back to Newton's law of gravitation and compute

\Delta U = \int_{R}^{R+h} \frac{GM_E m}{r^2}\;\text{d}r,

where R is the radius of the Earth, ME is the mass of the Earth, m is the mass of the man and h is the height of Everest.
 
Hootenanny said:
Yes. One would need to go back to Newton's law of gravitation and compute

\Delta U = \int_{R}^{R+h} \frac{GM_E m}{r^2}\;\text{d}r,

where R is the radius of the Earth, ME is the mass of the Earth, m is the mass of the man and h is the height of Everest.

Ah ok.

How would you go about integrating GMm/r^2? My maths isn't great but this is what I get:

Since G is a constant, factor it out.

\Delta U = G\int_{R}^{R+h} \frac{M_E m}{r^2}\;\text{d}r,

\Delta U = G [-\frac{M_E m}{r}]_{R}^{R+h}

Sorry, my maths isn't at this level yet!
 
studentxlol said:
Ah ok.

How would you go about integrating GMm/r^2? My maths isn't great but this is what I get:

Since G is a constant, factor it out.

\Delta U = G\int_{R}^{R+h} \frac{M_E m}{r^2}\;\text{d}r,

\Delta U = G [-\frac{M_E m}{r}]_{R}^{R+h}

Sorry, my maths isn't at this level yet!
So we are left with
\Delta U = GM_E m \left(\frac{1}{R} - \frac{1}{R+h}\right).
Simply plug the numbers in and you should get your result.
 

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