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I have learnt that change in potential energy in any conservative force field is equal to the negative of work done by that field through a distance. While applying the same for gravitational potential energy, I move a mass $m$ in the field of a large immovable mass $M$ from infinity to a point P which is a distance $x$ from the source mass, I obtain the work done as $$W = -\frac{GMm}{x}$$

But now, the change in potential energy will be negative of this work done and hence the it would be $$\Delta\text{PE (from infinity to P)} = -W = \frac{GMm}{x}$$

The problem is that this energy term is taken negative in the expression of total energy of a satellite. I don't know what I have done wrong?

Follow up question: while calculating total energy of the satellite in a orbit of radius r at a point p(say), we take potential energy term as the work done from infinity to that point but we have launched the satellite from earth surface to a height h (r=R+h, where R is the radius of earth) then why not take change in potential energy from surface to that point ?

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  • $\begingroup$ How do you get a negative work? Integrating $\int_{\infty}^x \vec{F} \cdot d\vec{s}$ gives a positive number, since $\vec{F}$ and $d\vec{s}$ both point towards the origin $\endgroup$
    – Prallax
    Commented Aug 14, 2021 at 5:50
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    $\begingroup$ Duplicate? I suspect that the problem lies with how you treated the incremental displacement to obtain W=-GMm/x. Have a look at the answers to Integrating a dot product gives wrong sign for work done and the links therein. $\endgroup$
    – Farcher
    Commented Aug 14, 2021 at 6:03
  • $\begingroup$ At the earth's surface the potential energy is even more negative than at height h. You can consider the change in potential energy from there, but you'll end up back at the same value at height h as taking it from infinity $\endgroup$
    – Thomas
    Commented Aug 15, 2021 at 9:42

2 Answers 2

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What you did wrong was to have a minus sign in your first equation, the work done should have been positive.

Imagine the satellite started at infinity and was stationary.

It has zero potential energy and no kinetic energy, so the total energy is zero.

If it then fell to the point $x$ near the mass $M$, the gravitational field has done positive work of value $$W = \frac{GMm}{x}$$ (the direction of the force and movement are the same).

This is potential energy lost by the gravitational field, so the field now has $W = -\frac{GMm}{x}$ stored in it. This energy is often regarded as part of the energy of the satellite

The energy has been converted to kinetic energy, so the total is still zero. .

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  • $\begingroup$ Thanks but this is not what i have asked $\endgroup$
    – Cyberax
    Commented Aug 14, 2021 at 8:11
  • $\begingroup$ @ Cyberax Ok, answer edited to make it clear where you went wrong. $\endgroup$ Commented Aug 14, 2021 at 8:14
  • $\begingroup$ Thanks this much is clear but one more thing i want to know if the work done from infinity to a point in field is equal to the work done from surface to to that point $\endgroup$
    – Cyberax
    Commented Aug 14, 2021 at 8:39
  • $\begingroup$ @ Cyberax usually they would be different, but you could find the work done from surface to to that point, by subtracting two GMm/x formulae, one with x for the point and one with the x for the surface. best of luck with it. $\endgroup$ Commented Aug 14, 2021 at 8:43
  • $\begingroup$ Hmm thanks that will do $\endgroup$
    – Cyberax
    Commented Aug 14, 2021 at 10:46
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There is nothing wrong. Your work is correct as far as I can see. The gravitational field does

  • negative work when an object is moved farther away, because then gravity is counteracting this displacement (e.g. sucking kinetic energy out), whereas it does

  • positive work when an object is moved closer, because gravity is helping (e.g. adding kinetic energy).

The sign that comes out from the math reflects this fact. No issues here.

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  • $\begingroup$ Ok i understand this much but then why potential energy term in total energy expression of satellite is always taken negative? $\endgroup$
    – Cyberax
    Commented Aug 14, 2021 at 8:42

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