"Gravitation" — Practice

Question 1

State the gravitational field and the gravitational potential at a point inside a uniform spherical shell of mass $M$ and radius $R$.

Accepted Answer

**Field: exactly zero** everywhere inside — not just at the centre.

**Potential: constant at $-\dfrac{GM}{R}$**, the same as its surface value.

These are consistent, not contradictory: $E=-dV/dr$, so zero field means the potential does not *change* inside. It does not mean the potential is zero.

Question 2

Escape velocity from Earth is $11.2\ 	ext{km/s}$. What is it for a body of twice the mass launched at $45°$ to the horizontal?

Accepted Answer

**Still $11.2\ 	ext{km/s}$.**

The mass cancels from both energy terms: $	frac12mv_e^{2}=GMm/R$.

The angle never enters, because the derivation uses only energy conservation and energy is a **scalar**. Provided nothing is in the way, direction is irrelevant.

Question 3

At what depth below the Earth's surface does $g$ fall to half its surface value?

Accepted Answer

$g_d=g\left(1-\dfrac{d}{R}ight)=\dfrac{g}{2}\Rightarrow\dfrac{d}{R}=\dfrac12$

$d=\dfrac{R}{2}\approx\mathbf{3200\ 	ext{km}}$

Note how different this is from the altitude case — you would have to go far further *up* to halve $g$, because that falls as $1/r^{2}$ rather than linearly.

Question 4

Four particles of mass $m$ sit at the corners of a square of side $a$. Find the gravitational field and potential at its centre.

Accepted Answer

**Field: zero.** Diagonally opposite masses pull equally in opposite directions and cancel in pairs.

**Potential:** each corner is $\dfrac{a}{\sqrt2}$ from the centre, and potentials add as scalars:

$V=4	imes\left(-\dfrac{Gm}{a/\sqrt2}ight)=\mathbf{-\dfrac{4\sqrt2\,Gm}{a}}$

Non-zero potential with zero field — the same pairing that made the field cancel does nothing to the potential, because scalars have no direction to oppose.

Question 5

At what height does $g$ fall to 81% of its surface value? Take $R=6400\ 	ext{km}$.

Accepted Answer

Exact form: $\dfrac{g_h}{g}=\left(\dfrac{R}{R+h}ight)^{2}=0.81\Rightarrow\dfrac{R}{R+h}=0.9$

$h=R\left(\dfrac{1}{0.9}-1ight)=\dfrac{R}{9}\approx\mathbf{711\ 	ext{km}}$

The approximation $g(1-2h/R)=0.81g$ would give $h=0.095R\approx640$ km — **over 10% out**, because $h$ is no longer small compared with $R$.

Question 6

A planet has half the Earth's radius and the same mean density. Compare its escape velocity with Earth's.

Accepted Answer

Same density $\Rightarrow M=ho\cdot	frac43\pi R^{3}\propto R^{3}$.

$v_e=\sqrt{\dfrac{2GM}{R}}\propto\sqrt{\dfrac{R^{3}}{R}}=R$

So $v_e$ is **half** Earth's, about $\mathbf{5.6\ 	ext{km/s}}$.

Not $1/\sqrt2$ times — that would be the answer at fixed *mass*. Substituting for $M$ before scaling is the whole question.

Question 7

A moon orbits a planet at radius $r$ with period $T$. A second moon of the same planet has period $8T$. Find its orbital radius.

Accepted Answer

$T^{2}\propto r^{3}\Rightarrow r\propto T^{2/3}$

$\dfrac{r_2}{r}=8^{2/3}=4$, so $r_2=\mathbf{4r}$.

The moons' masses never enter — the constant $4\pi^{2}/GM$ depends only on the **central** body. That is exactly what makes Kepler's third law a way of weighing planets and stars.

Question 8

At what depth is $g$ equal to its value at a height $h$ above the surface, for $h\ll R$?

Accepted Answer

$g\left(1-\dfrac{2h}{R}\right)=g\left(1-\dfrac{d}{R}\right)$

$\Rightarrow \mathbf{d=2h}$

The depth is **twice** the height, because the altitude correction carries a factor of 2 (from the inverse square) while the depth correction does not (it is linear).

Question 9

A satellite of mass 500 kg orbits at radius $2R$. Find the energy needed to raise it to $4R$, and state what happens to its speed. Take $g=10$, $R=6400\ 	ext{km}$.

Accepted Answer

$E=-\dfrac{GMm}{2r}$, so

$\Delta E=-\dfrac{GMm}{8R}+\dfrac{GMm}{4R}=\dfrac{GMm}{8R}$

Substituting $GM=gR^{2}$: $\Delta E=\dfrac{gRm}{8}=\dfrac{10(6.4	imes10^{6})(500)}{8}=\mathbf{4.0	imes10^{9}\ 	ext{J}}$

**The speed falls** by a factor of $\sqrt2$, since $v_o\propto1/\sqrt{r}$. Energy up, speed down — the signature result of the chapter.

Question 10

Explain why the gravitational field inside a uniform spherical shell is exactly zero, and why this would fail for an inverse-cube force law.

Accepted Answer

Take any interior point and draw a narrow **double cone** through it, cutting two patches of shell on opposite sides.

The nearer patch pulls more strongly, by a factor $1/r^{2}$. But it also subtends a smaller area, by exactly $r^{2}$ — the area of a cone's cross-section grows as the square of the distance.

The two factors **cancel exactly**, so the two patches pull equally and oppositely. Every double cone through the point does the same, so the total is zero.

**For an inverse-cube law** the pull would scale as $1/r^{3}$ while the area still scales as $r^{2}$, leaving a net $1/r$ imbalance toward the nearer wall. A residual field would survive.

This makes the null result one of the sharpest experimental tests of the exponent being exactly 2.

Question 11

Four particles, each of mass $m$, sit at the corners of a square of side $a$. Find the total gravitational potential energy of the system.

Accepted Answer

Four particles give $\dfrac{4(3)}{2}=6$ pairs — but they are of two kinds:

| Pairs | How many | Separation |
|---|---|---|
| Along the sides | 4 | $a$ |
| Along the diagonals | 2 | $a\sqrt2$ |

$U=-Gm^{2}\left[\dfrac{4}{a}+\dfrac{2}{a\sqrt2}ight]=-\dfrac{Gm^{2}}{a}\left[4+\sqrt2ight]$

$$U=-\dfrac{5.41\,Gm^{2}}{a}$$

*The two standard errors:* counting 12 pairs by treating AB and BA as distinct, and forgetting that the diagonals are longer by $\sqrt2$ so they contribute less.

The work required to separate the four to infinity is $+5.41\,Gm^{2}/a$.

Question 12

A $1000\ 	ext{kg}$ satellite is in a circular orbit of radius $2R$, where $R=6400\ 	ext{km}$ and surface $g=9.8$. Find the extra energy needed to make it just escape, and the factor by which its kinetic energy must rise.

Accepted Answer

$GM=gR^{2}=9.8(6.4	imes10^{6})^{2}=4.01	imes10^{14}$, and $r=1.28	imes10^{7}\ 	ext{m}$.

**Orbital kinetic energy:**

$K_o=\dfrac{GMm}{2r}=\dfrac{4.01	imes10^{14}(1000)}{2(1.28	imes10^{7})}=1.57	imes10^{10}\ 	ext{J}$

**Total energy** $E=-K_o=-1.57	imes10^{10}\ 	ext{J}$, so the extra energy needed to reach $E=0$ is

$$\Delta E=\mathbf{1.57	imes10^{10}\ 	ext{J}}$$

**The kinetic energy exactly doubles.** Since $v_e=\sqrt2\,v_o$ at any radius, $K_e=2K_o$ — and the extra kinetic energy required, $K_o$, is precisely the binding energy.

*That coincidence is worth keeping:* for any circular orbit, the energy needed to escape equals the kinetic energy the satellite already has.

Practice — "Gravitation"

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