- #1

Ebby

- 41

- 14

- Homework Statement
- What is the work done by gravity in moving the particle from a distance of infinity to a distance R from the centre of the Earth (where R > the radius of the Earth)?

- Relevant Equations
- W = F * r

What is the work done by gravity in moving the particle from a distance of ##\infty## to a distance ##R## from the centre of the Earth (where ##R## > the radius of the Earth)?

The answer is obvious, since the displacement and the force of gravity are in the same direction. Therefore, gravity does positive work in the amount ##W_{grav} = GMm/R##.

However, I want to show you how I am tempted to formulate my answer wrongly, which leads to an answer of the wrong sign. I am hoping someone can explain clearly why this is wrong. I keep coming back to problems like this, and am plagued by the same confusion every time. I am hoping finally to resolve this issue in my mind!

So, we're going to use the equation generally given with the form: ##W = \int_a^b \vec F \cdot \, d\vec r##. In this particular case we have: $$W_{grav} = \int_\infty^R -\frac {GMm} {x^2} \, {\hat x} \cdot -dx \, \hat x$$

$$= \int_\infty^R \frac {GMm} {x^2} \, dx$$

$$= GMm \int_\infty^R \frac 1 {x^2} \, dx$$

$$= \frac {-GMm} {x} \left. \right |_{\infty}^R$$

$$= \frac {-GMm} {R}$$

Which is of course wrong.

Additionally, I have difficulty thinking of a force (gravity in this case) doing positive work on an object, and yet the object having less P.E. afterwards, not more. It's more intuitive for me to think of an object being "worked on" and consequently having more energy. What is a better way to think about this?

EDIT: I see that my sketch is slightly wrong. There's a ##\hat x## that needs to be the numerator, not the denominator. Anyway, all the LaTeX is fine.

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