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To be specific, I'm asking why does the mass need to be moved "at a constant speed"?

My textbook says it is so that no kinetic energy is involved. But shouldn't kinetic energy be involved in order to move the point mass from infinity to a point in a gravitational field? Since kinetic energy is converted to gravitational potential energy.

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We do this so gravitational potential energy is easier to calculate. Saying the speed is constant implies we are applying an equal and opposite force to that of gravity throughout the process (we have an equation for this). This has the added benefit of guarantying we don't overestimate the energy needed to escape the gravitational pull. We could apply a force that varies from the gravitational force. We just have to be careful not to overdo it.

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Gravitational potential has a "floating" value, something like voltage. Any arbitrary point can be chosen and its gravitational potential can be called "zero". The gravitational potential at any other point in the universe is then the difference in gravitational potential between the chosen point and the other point. The same principle applies to electrical potential.

"Potential" is different from "potential energy". Potential is the energy (difference) per unit mass or unit charge depending on whether we're talking about gravity or electromagnetism. Potential energy of a mass is the gravitational potential multiplied by the mass; potential energy of a charge is the electrical potential multiplied by the charge.

To measure gravitational potential at a point X, we can (in principle) place a tiny mass at the "zero" point and let it fall to X. At X we measure the kinetic energy of the tiny mass. All of the kinetic energy came from the change in gravitational potential energy, so we divide the kinetic energy by the tiny mass, and that is the negative of the change in gravitational potential energy.

An alternate way to measure gravitational potential at X is to move the tiny mass slowly from the zero point to point X, and measure the direction and magnitude of force that must be exerted on the mass all along its path. The net work done by the mass in getting from "zero" to X is the integral of the dot product of the force with the unit tangent vector to the path, all along the length of the path. That work is equal to the nega change in gravitational potential energy, so the change in gravitational potential is obtained by dividing by the mass. This can be a complicated measurement. However, if the measurement is done, the result is the same as the first method.

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If you consider a generic stationary mass distribution, it generates a gravitational stationary field in all the space around: $$\mathbf G=\mathbf G(x,y,z)$$ You can demonstrate that always exists a function called potential (not potential energy) for which: $$grad(V(x,y,z))=\mathbf G(x,y,z)$$ in every point of the space. Also you can demonstrate that: $$\int_{\beta} \mathbf G \cdot \mathbf {dr} =V(\mathbf b)-V(\mathbf a)$$ Where $\beta$ is a line which starts in $\mathbf a$ and ends in $\mathbf b$. You can obtain also that: $$\int_{\beta} m\mathbf G \cdot \mathbf {dr} =mV(\mathbf b)-mV(\mathbf a)$$ If you define $$U=Vm$$ the potential energy own by a mass, finaly you have: $$W_{\beta}=\int_{\beta} \mathbf F \cdot \mathbf {dr} =U(\mathbf b)-U(\mathbf a)=\Delta k_{a-b}$$ However the kinetic energy is not involved in the defintion of the potential energy.

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