Missing mass in gravitational force

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The discussion clarifies the relationship between gravitational potential and gravitational force. It emphasizes that the formula \(\vec{F} = -\nabla \phi\) applies to a unit test mass, distinguishing potential \(\phi\) from potential energy \(U\). The gravitational force is derived from the gradient of the potential, leading to the expression for the gravitational field. The importance of checking units is highlighted to avoid confusion between potential and potential energy. Understanding these concepts is crucial for correctly applying the principles of conservative forces in physics.
PLuz
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This is a really dumb question but I can't seem to make sense out of this...

For a conservative force, we have \vec{F}=-\nabla \phi, where \phi stands for the potential. So let's take the gravitational potential, given by:
\phi=-G_N \frac{M}{r}.
Then, by the previous formula: \vec{F_g}=-G_N \frac{M}{r^2}\hat{e_r}...but this is the expression for the gravitational field (force per unit mass) not the gravitational force... What am I doing wrong?


Thank you very much.
 
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No, ##a = -\nabla \phi##. ##F = -\nabla \phi## holds for a unit test mass. You are confusing the potential ##\phi## with the potential energy ##U## for which it is true that ##F = -\nabla U##. Whenever you have doubts like this, just check units. ##\phi## has units of Joules/kg so ##\nabla \phi## has units of Joules/(kg*m) = N*m/(kg*m) = N/kg = m/s^2
 
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