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As the height from Earth increases, the gravitational potential also increases. If I look at this from the definition of potential (work done in bringing a unit mass from infinity to that point) the potential value would be high for lower distances (from infinity), so does this mean that the Earth will pull the farther objects with more force?

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The relation between force and potential energy is, in general, $$ {\bf F}=-\nabla U(r) $$ Using spherical coordinates, the unique component of the force (purely radial) is $$ F_r=-\frac{{\mathrm d}U}{{\mathrm d}r}. $$ Therefore, the value of the potential energy does not matter. What matters is its slope. Wherever is the zero of the potential energy, the slope of $U(r)$ decreases with the distance. Forces at larger distances are weaker than forces at smaller distances.

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As the height from Earth increases, the gravitational potential also increases.

This is not quite true.

First, Let's not make the approximation of constant $g$ which is valid for small distances from the surface of the earth. $$dU=-\int \mathbf{F}\cdot d\mathbf{s}=\int \frac{k}{r^2}dr$$ $$U(r_2)-U(r_1)=k\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$ Setting $U(r_\text{surface})=0$ that is surface of earth is zero potential. $$U(r)=\frac{k}{r}$$ $$\lim_{r\rightarrow \infty}U(r)=0$$ As expected. The gravitational potential decreases as height increase if the surface is taken to zero potential.


If you make approximation that $g$ is constant then $$dU=-\int -mg\hat{r}\cdot d\mathbf{s}=mg(r_2-r_1)$$ Then it look like $U(r)\propto r$ but that is not true.

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  • $\begingroup$ Wrong answer. If you write explicitly the limits of integration in the integrals, you'll realize that even putting zero the potential energy at the surface of Earth, the potential increases with the distance. If it would decrease, taking into account the relation between force and potential (${\bf F}=-\nabla U$), there should be a repulsive force. $\endgroup$ Jul 1, 2021 at 14:59

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