Exercises — Gravitational field intensity g = GM - r²

(a) $g$ is the gravitational force per unit mass that a small test mass would feel at that point — a property of the location, not of the visitor. (b) The force on the test mass is $F=GMm/r^2$, which is proportional to $m$. Field is $g=F/m$, so dividing by $m$ cancels it: $g=GM/r^2$. What remains depends only on the source $M$ and the distance $r$.

Use $r=R_{\text{moon}}=1.74\times 10^6\text{m}$ (distance from the Moon's centre): $g=\frac{G{M}_{\text{moon}}}{R_{\text{moon}}^2}=\frac{(6.67\times 10^{-11})(7.35\times 10^{22})}{(1.74\times 10^6{)}^2}$ Numerator: $6.67\times 7.35=49.0$, so $49.0\times 10^{11}=4.90\times 10^{12}$. Denominator: $(1.74{)}^2=3.03$, so $3.03\times 10^{12}$. $g=\frac{4.90\times 10^{12}}{3.03\times 10^{12}}\approx 1.62{\text{N kg}}^{-1}$ That is about $\frac{1}{6}$ of Earth's $9.8$ — why astronauts bounce.

By the definition $\vec{g}=\vec{F}/m$, rearranged $F=mg$.

• $2\text{kg}$: $F=2\times 9.8=19.6\text{N}$.
• $50\text{kg}$: $F=50\times 9.8=490\text{N}$. The field $g$ is the same $9.8$ for both — only the force scales with the visitor's mass.

Altitude $h=2R_{\oplus }$ means distance from centre $r=R_{\oplus }+h=3R_{\oplus }$. Let $g_0$ stand for the surface gravity of Earth — the field at $r=R_{\oplus }$, which we know is $g_0=9.8{\text{N kg}}^{-1}$. We use it as a reference so we can scale by a ratio instead of re-plugging $G$ and $M$. Field scales as $1/r^2$, so compared to that surface value $g_0$: $g=g_0{\left(\frac{R_{\oplus }}{3R_{\oplus }}\right)}^2=g_0\cdot \frac{1}{9}=\frac{9.8}{9}\approx 1.09{\text{N kg}}^{-1}$ Why $3R$, not $2R$? Altitude is measured from the surface; $r$ in the formula is from the centre, so add $R_{\oplus }$.

Set $g=GM/r^2=1.0$ and solve for $r$: $r=\sqrt{\frac{G{M}_{\oplus }}{g}}=\sqrt{\frac{(6.67\times 10^{-11})(5.97\times 10^{24})}{1.0}}$ Numerator: $6.67\times 5.97=39.8$, so $GM=3.98\times 10^{14}$. $r=\sqrt{3.98\times 10^{14}}=\sqrt{39.8\times 10^{13}}\approx 1.996\times 10^7\text{m}$ That is about $2.0\times 10^7\text{m}$, roughly $3.13R_{\oplus }$ from the centre.

$\frac{g_{\text{planet}}}{g_{\oplus }}=\frac{G{M}_p/R_p^2}{G{M}_{\oplus }/R_{\oplus }^2}=\frac{M_p}{M_{\oplus }}\cdot {\left(\frac{R_{\oplus }}{R_p}\right)}^2=(2)\cdot {\left(\frac{1}{2}\right)}^2=2\cdot \frac{1}{4}=\frac{1}{2}$ So $g_{\text{planet}}=\frac{1}{2}{g}_{\oplus }\approx 4.9{\text{N kg}}^{-1}$. The mass doubles the field but the radius (squared, in the denominator) quarters it — the radius wins.

In the figure, the two amber dots are the equal masses $M$ set $2a$ apart; the small cyan dot at the centre is our test point, a distance $a$ from each mass (the two white "$a$" brackets underneath confirm this). Each mass alone produces a field of magnitude $g_{\text{each}}=\frac{GM}{a^2}$ pointing toward its own mass. The two cyan arrows in the figure show these contributions: the left one points left (toward the left mass) and the right one points right (toward the right mass) — exactly opposite directions. $g_{\text{net}}=g_{\text{each}}-g_{\text{each}}=0$ The midpoint is a null point: the fields cancel by symmetry. A test mass placed exactly there feels no net gravitational pull (though it is an unstable balance — nudge it and it falls toward the nearer mass).

In the figure, the large amber dot on the left is $4M$ and the smaller dot on the right is $M$; the cyan dot marks the null point, with the two cyan arrows showing $4M$'s field (pulling left) and $M$'s field (pulling right) opposing each other there. The white brackets label the distances $x$ (from $4M$) and $d-x$ (from $M$). The null point must lie between them (both fields point inward toward their own mass, so only between them can they oppose and cancel). Let it be distance $x$ from $4M$, hence $d-x$ from $M$. Set magnitudes equal: $\frac{G(4M)}{x^2}=\frac{GM}{(d-x{)}^2}$ Cancel $GM$: $\frac{4}{x^2}=\frac{1}{(d-x{)}^2}\Rightarrow 4(d-x{)}^2=x^2$ Take the positive square root of both sides ($x$ and $d-x$ are both positive lengths): $2(d-x)=x\Rightarrow 2d-2x=x\Rightarrow 2d=3x\Rightarrow x=\frac{2d}{3}$ The null point sits at $x=\frac{2}{3}d$ from the heavy mass — closer to the lighter mass (as the figure shows, the cyan dot is nearer to $M$), which makes sense: you must stand nearer to $M$ so its weaker field has a short enough distance to keep up with the strong field of $4M$. Why reject the other root? Taking the negative root gives $-2(d-x)=x\Rightarrow x=2d$, which is outside the segment where both fields point the same way — no cancellation there.

In the figure, the horizontal cyan arrow is $g_E$ (East) and the vertical cyan arrow is $g_M$ (North); the amber arrow is the resultant $g_{\text{net}}$, the diagonal of the dashed rectangle. Because $g_E$ and $g_M$ meet at a right angle, they form the two legs of a right triangle whose hypotenuse is $g_{\text{net}}$ — this is why we combine them with Pythagoras, not by adding numbers: $g_{\text{net}}=\sqrt{g_E^2+g_M^2}=\sqrt{(3.0{)}^2+(4.0{)}^2}\times 10^{-3}=\sqrt{9+16}\times 10^{-3}=\sqrt{25}\times 10^{-3}$ $g_{\text{net}}=5.0\times 10^{-3}{\text{N kg}}^{-1}$ Direction, measured North of East (the amber arc in the figure): $\theta =\arctan \left(\frac{g_M}{g_E}\right)=\arctan \left(\frac{4.0}{3.0}\right)=\arctan (1.333)\approx 53.1^{\circ }$ Why $\arctan$? The tangent of the angle is opposite over adjacent $=g_M/g_E$; $\arctan$ answers "which angle has this tangent?" Both components are positive (East and North), so the answer sits in the first quadrant — no sign correction needed.

By Newton's Second Law (F=ma), acceleration $a=F/m=g$ (the mass cancels — this is the same cancellation as in the field definition). So the acceleration is simply $a=g=1.62{\text{m s}}^{-2}$ independent of the $10\text{kg}$. Starting from rest, $v=at$: $v=1.62\times 3.0=4.86{\text{m s}}^{-1}$ Why is $a$ numerically equal to $g$? Because $g=F/m$ (field) and $a=F/m$ (Newton's 2nd law) are the same $F/m$. The units ${\text{N kg}}^{-1}$ and ${\text{m s}}^{-2}$ are therefore identical.

This connects to Gravitational Potential and Potential Energy. Differentiate $V=-GM{r}^{-1}$ with respect to $r$: $\frac{dV}{dr}=-GM\cdot (-1)r^{-2}=\frac{GM}{r^2}$ Then $g=-\frac{dV}{dr}=-\frac{GM}{r^2}$ The magnitude is $GM/r^2$ (matching the parent formula ✔). The minus sign says $g$ points in the direction of decreasing $r$ — i.e. inward, toward $M$. The field is the slope of the potential landscape, and things roll downhill toward the source.

See Electric Field Intensity E=kQ/r². The two laws are algebraically identical inverse-square fields: $g=\frac{GM}{r^2},E=\frac{kQ}{r^2}$

Match ::: role $G\leftrightarrow k$ ::: the coupling constant of the interaction $M\leftrightarrow Q$ ::: the "source" (mass vs charge) $g=F/m\leftrightarrow E=F/q$ ::: force per unit test property (mass vs charge)

Key physical difference: mass is always positive, so $g$ only attracts (always inward). Charge can be $+$ or $-$, so $E$ can point inward or outward. Both dilute as $1/r^2$ for the same geometric reason — field lines spreading over a sphere of area $4\pi r^2$, formalised by Gauss's Law for Gravity.

(a) Let the null point be distance $x$ from the lighter mass $M$, so $d-x$ from $9M$, with $d=1.0\times 10^8$. Set field magnitudes equal: $\frac{GM}{x^2}=\frac{G(9M)}{(d-x{)}^2}\Rightarrow \frac{1}{x^2}=\frac{9}{(d-x{)}^2}$ $(d-x{)}^2=9x^2\Rightarrow d-x=3x\Rightarrow d=4x\Rightarrow x=\frac{d}{4}=2.5\times 10^7\text{m}$ So the null point is $2.5\times 10^7\text{m}$ from $M$ (and $7.5\times 10^7\text{m}$ from $9M$) — closer to the smaller mass, as expected. (b) Nudge it toward $9M$: now it is closer to $9M$, so $9M$'s field on it grows while $M$'s field shrinks — the net pull is toward $9M$, pulling it further away from the null point. It runs away (does not return). The null point is an unstable equilibrium — a saddle in the potential landscape.

At the centre: every tiny chunk of the sphere has an identical chunk directly opposite it, at the same distance. Their two field contributions point in exactly opposite directions and cancel in pairs. Summing over all such pairs gives $g=0$ at the centre — by pure symmetry. Inside ($r<R$): by the shell theorem (a consequence of Gauss's Law for Gravity), only the mass inside radius $r$ contributes — call it the enclosed mass $M_{\text{enc}}$, the portion of the sphere lying closer to the centre than our point. For a uniform sphere (constant density $\rho$), this enclosed mass is $M_{\text{enc}}=\rho \cdot \frac{4}{3}\pi r^3$, so it grows as $r^3$ (whereas the total mass $M=\rho \cdot \frac{4}{3}\pi R^3$ is the full-sphere value at $r=R$). Then $g_{\text{inside}}=\frac{G{M}_{\text{enc}}}{r^2}\propto \frac{r^3}{r^2}=r$ — $g$ rises linearly from $0$ at the centre. At the surface ($r=R$): now $M_{\text{enc}}=M$ (the whole mass), and $g$ reaches its maximum, $g_{\max }=GM/R^2$. Outside ($r>R$): the full mass $M$ acts as if concentrated at the centre, so $g=GM/r^2$, which decreases as $1/r^2$ — the ordinary inverse-square falloff. In short: $g$ climbs linearly from $0$ at the centre to $GM/R^2$ at the surface, then decays as $1/r^2$ beyond it.

No. Potential and field are different beasts:

• Field $\vec{g}$ is a vector; opposing fields can cancel to zero.
• Potential $V=-\frac{GM}{r}$ is a scalar and is always negative for mass; two negatives cannot cancel — they add to a more negative number. At the null point, using $x=2.5\times 10^7$ and $d-x=7.5\times 10^7$: $V=-\frac{GM}{x}-\frac{G(9M)}{d-x}=-GM\left(\frac{1}{2.5\times 10^7}+\frac{9}{7.5\times 10^7}\right)$ $=-GM\left(4.0\times 10^{-8}+1.2\times 10^{-7}\right)=-GM(1.6\times 10^{-7}){\text{J kg}}^{-1}$ which is decidedly not zero (it is negative). The field being zero means "no net pull here"; the potential being negative means "you'd still have to do work to escape from here." Zero force $\ne$ zero energy.