Gravitational force

Newton's law of universal gravitation: any two masses attract each other with a force $\vec{F}_g = -\dfrac{Gm_1 m_2}{r^2}\hat{r}$, where $r$ is the distance between their centers and $\hat{r}$ points from the source toward the object. The minus sign makes the force attractive, pointing back toward the source. The constant $G = 6.67\times10^{-11}$ N·m$^2$/kg$^2$ is the same everywhere in the universe.

Split the force into a source part and a test-object part. The gravitational field of a mass $M$ is $\vec{g} = \dfrac{\vec{F}_g}{m} = -\dfrac{GM}{r^2}\hat{r}$: the force per kilogram a small test mass would feel. Its magnitude $g = \dfrac{GM}{r^2}$ is in N/kg, the same number as the free-fall acceleration in m/s$^2$ whenever gravity is the only force.

Near a planet's surface, this field is just little-$g$. At Earth's surface, $r$ is one Earth radius from the center and $g = 10$ N/kg. That one value causes most of the trouble here: it is a single point on the $GM/r^2$ curve, not a fixed constant.

Two ideas run through the rest of the topic. The field weakens with distance as $1/r^2$ (§2), and the rule changes completely once you go inside a body (§5).

The field is $g = \dfrac{GM}{r^2}$, so it falls off as the inverse square of distance from the center. Double the distance and the field drops to a quarter; triple it and it drops to a ninth. The surface value $10$ N/kg holds only at $r = R_E$.

Ratios make this easy. Write $\dfrac{g(r)}{g_{\text{surface}}} = \left(\dfrac{R_E}{r}\right)^2$. At $r = 2R_E$ the ratio is $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$, so $g = 10/4 = 2.5$ N/kg. At $r = 3R_E$ it is $\tfrac{1}{9}$, so $g = 10/9 \approx 1.1$ N/kg. No need to plug in $G$ or $M$ at all.

The same logic works across planets. Surface field is $g = \dfrac{GM}{R^2}$, so surface gravity scales as mass over radius squared: $\dfrac{g_{\text{planet}}}{g_{\text{Earth}}} = \dfrac{M_{\text{planet}}/M_E}{(R_{\text{planet}}/R_E)^2}$. A planet with the same mass but twice the radius has a quarter of Earth's surface gravity.

The trap is carrying $10$ everywhere: into orbit, onto the Moon, out to a distant satellite. The number $10$ answers one specific question, the field at Earth's surface. Every other distance or planet needs $g = GM/r^2$, most easily as a ratio.

Setup. (a) A satellite orbits at $r = 4R_E$ from Earth's center; Earth's surface field is $10$ N/kg. Find the field there. (b) Planet Z has $3$ times Earth's mass and twice its radius. Find its surface field.

(a) Use the distance ratio. The field falls as $1/r^2$: $\dfrac{g}{g_{\text{surface}}} = \left(\dfrac{R_E}{4R_E}\right)^2 = \dfrac{1}{16}$, so $g = 10/16 = 0.625$ N/kg.

(b) Scale mass and radius. Surface field scales as $M/R^2$: $\dfrac{g_Z}{g_E} = \dfrac{3}{2^2} = \dfrac{3}{4}$, so $g_Z = 10\left(\tfrac{3}{4}\right) = 7.5$ N/kg.

The trap to avoid. Neither answer is $10$. Far out, the field is a small fraction of the surface value; on Planet Z, the larger radius outweighs the larger mass. Reusing $10$ ignores both the distance and the $GM/R^2$ scaling.

True weight is the gravitational force, $W = mg$, and it changes only when $g$ changes (a different distance or planet). Apparent weight is different: it is the normal force a scale or floor pushes back with, the number a scale actually shows.

In this scale setup, where the only vertical forces are the normal force $N$ and gravity $mg$, the two are equal only when there is no vertical acceleration. Apply the second law along the vertical for a person on a scale: $N - mg = ma$, so $N = m(g + a)$. Standing still, $a = 0$ and $N = mg$, the scale reads the true weight. Accelerating upward, $N > mg$; accelerating downward, $N < mg$.

Take the extreme case. In free fall the only force is gravity, so $a = -g$ and $N = m(g - g) = 0$: the scale reads zero. That is weightlessness, not the absence of gravity but the absence of a support force. An astronaut in orbit is in continuous free fall toward Earth, with gravity fully present (at $r = 2R_E$ it is still $2.5$ N/kg), yet floats because nothing pushes up.

The trap is reporting $mg$ when the question asks what the scale reads, or concluding that "weightless" means gravity has switched off. Apparent weight is the normal force; solve for it from the free-body diagram. In orbit, gravity is still there; what is missing is the floor.

So far the source was a point, or we stood outside a sphere. Inside a body the rule changes, and Newton's shell theorem tells us how. For a thin spherical shell of mass $M$ and radius $R$: a test mass outside ($r > R$) feels the shell as if all of $M$ sat as a point at the center, $F = \dfrac{GMm}{r^2}$; a test mass inside ($r < R$) feels exactly zero net force, everywhere inside, because the pulls from all parts of the shell cancel.

Swapping these two rules is the classic mistake: writing $GMm/r^2$ for a point inside the shell, or declaring zero force for a point outside. Pick the rule by comparing $r$ to $R$. Inside means zero; outside means point mass.

A solid sphere is a stack of nested shells. For a test mass at radius $r$ inside a uniform solid sphere, every shell outside $r$ has the test mass inside it and so contributes zero; only the material within $r$ pulls, acting as a point mass at the center. The mass that matters is the partial mass within $r$, set by the volume ratio: $m_{\text{within}} = M\left(\dfrac{r}{R}\right)^3$.

Put that into the inverse-square law: $F = \dfrac{G\,m_{\text{within}}\,m}{r^2} = \dfrac{GMm}{R^3}\,r$. The $r^3$ on top beats the $r^2$ below, so the force grows in proportion to $r$ inside the sphere: zero at the center, largest at the surface, then $1/r^2$ beyond. The trap is reaching for the outside formula (full $M$, $1/r^2$) at an inside point, which both uses too much mass and gets the dependence backwards.

Setup. A small mass $m$ sits at $r = R/2$ inside a solid uniform sphere of total mass $M$ and radius $R$. Find the gravitational force on it, and compare with the (wrong) outside formula.

Find the partial mass. Only the mass within $r = R/2$ acts. By volume, $m_{\text{within}} = M\left(\dfrac{r}{R}\right)^3 = M\left(\dfrac{1}{2}\right)^3 = \dfrac{M}{8}$.

Apply the inverse-square law with that mass. $F = \dfrac{G(M/8)m}{(R/2)^2} = \dfrac{G(M/8)m}{R^2/4} = \dfrac{GMm}{2R^2}$. Equivalently, $F = \dfrac{GMm}{R^3}r$, which at $r = R/2$ gives the same $\dfrac{GMm}{2R^2}$.

The trap to avoid. The outside formula would give $F = \dfrac{GMm}{(R/2)^2} = \dfrac{4GMm}{R^2}$, eight times too large. It uses the full mass $M$ when only $M/8$ is enclosed, and treats the field as $1/r^2$ when inside it grows as $r$. Inside a solid sphere, count only the partial mass and remember the force rises with $r$.