Chapter 7: Gravitation (Gravitation)

(a) No. Gravitational force does not depend on the medium, and a hollow shell does not stop the pull of outside matter on a body inside. Gravitational shielding is not possible.

(b) Yes, if the station is large enough. In a small ship everything falls together, so gravity cannot be felt. In a very large station the value and direction of g differ measurably from one end to the other, and this variation (tidal effect) can be detected.

(c) The tidal effect depends on the difference of the pull across the earth's diameter, which varies as $\frac{1}{r^3}$, whereas the force varies as $\frac{1}{r^2}$. The moon is much closer than the sun, so its tidal effect is larger even though its pull is smaller.

(a) No (b) Yes, if the station is large enough (c) Tidal effect ∝ 1/r³, and the moon is much closer.

(a) $g(h)=\frac{GM}{(R+h{)}^2}$: decreases. (b) $g(d)=g\left(1-\frac{d}{R}\right)$: decreases. (c) $g=\frac{GM}{R^2}$ does not contain the body's mass: independent of the mass of the body. (d) $mg(r_2-r_1)$ assumes g constant, which is only an approximation; the first formula is exact: more accurate.

(a) decreases (b) decreases (c) mass of the body (d) more

"Twice as fast" means half the period: $T_p=\frac{1}{2}{T}_e$. By Kepler's third law, $r^3\propto T^2$:

$\frac{r_p}{r_e}={\left(\frac{T_p}{T_e}\right)}^{2/3}={\left(\frac{1}{2}\right)}^{2/3}=0.63$.

0.63 times the earth's orbit (smaller by a factor of 0.63)

$M=\frac{4{\pi }^2{r}^3}{G{T}^2}$, so for Jupiter (Io) and the sun (earth: $r=1.496\times 10^{11}$ m, $T=365.25$ days):

$$\frac{M_J}{M_S}={\left(\frac{r_{Io}}{r_E}\right)}^3{\left(\frac{T_E}{T_{Io}}\right)}^2={\left(\frac{4.22\times 10^8}{1.496\times 10^{11}}\right)}^3{\left(\frac{365.25}{1.769}\right)}^2=(2.82\times 10^{-3}{)}^3\times (206.5{)}^2$$

$=2.25\times 10^{-8}\times 4.26\times 10^4=9.6\times 10^{-4}\approx \frac{1}{1000}$.

M_J ≈ M_S/1000. Shown.

$M=2.5\times 10^{11}\times 2\times 10^{30}=5\times 10^{41}$ kg; $r=50000\times 9.46\times 10^{15}=4.73\times 10^{20}$ m.

$$T=2\pi \sqrt{\frac{r^3}{GM}}=2\pi \sqrt{\frac{(4.73\times 10^{20}{)}^3}{6.67\times 10^{-11}\times 5\times 10^{41}}}=2\pi \sqrt{3.17\times 10^{30}}=1.12\times 10^{16}\text{s}$$

In years: $\frac{1.12\times 10^{16}}{3.15\times 10^7}$.

About 3.5 × 10⁸ years

(a) $K=\frac{GMm}{2r}$, $E=-\frac{GMm}{2r}=-K$: kinetic energy.

(b) The satellite already has kinetic energy, so it needs only $\frac{GMm}{2r}$, while a stationary object needs $\frac{GMm}{r}$: less.

(a) kinetic (b) less

$v_e=\sqrt{\frac{2GM}{R+h}}$: it depends on the gravitational potential at the launch point, not on the body's mass or the direction.

(a) No (b) No (c) No (d) Yes (it depends on the potential at the launch point, so slightly on height and latitude)

Gravity is a central force (no torque about the sun), so angular momentum is constant; it is conservative, so total energy is constant. The distance from the sun changes, so speed, angular speed, KE and PE all change.

Only (c) angular momentum and (f) total energy are constant.

In weightlessness, body fluids are no longer pulled towards the feet; they shift upwards, swelling the face and causing headache. Without a sense of "up", orientation is lost. The feet do not swell.

(b), (c) and (d)

Complete the hemisphere into a full shell: inside a full shell the potential is constant and the intensity is zero. The upper (missing) half and the lower half must therefore give equal and opposite intensities at C. The lower hemisphere pulls straight down by symmetry.

(iii) c

The same argument holds at any point P on the flat face: the intensities of the two halves are equal and opposite, and the upper half's field at P is symmetric about the vertical, so the lower half's field at P is vertical, pointing down.

(ii) e

At distance x from the earth: $\frac{G{M}_E}{x^2}=\frac{G{M}_S}{(r-x{)}^2}\Rightarrow \frac{x}{r-x}=\sqrt{\frac{M_E}{M_S}}=\sqrt{3\times 10^{-6}}=1.732\times 10^{-3}$.

$x=\frac{1.732\times 10^{-3}}{1.001732}\times 1.5\times 10^{11}=2.6\times 10^8$ m.

2.6 × 10⁸ m from the earth's centre

The sun's gravity provides the earth's centripetal force: $\frac{G{M}_{S}m}{r^2}=m\frac{4{\pi }^2}{T^2}r\Rightarrow M_S=\frac{4{\pi }^2{r}^3}{G{T}^2}$.

With $r=1.5\times 10^{11}$ m, $T=365\times 24\times 3600=3.15\times 10^7$ s:

$$M_S=\frac{39.5\times (1.5\times 10^{11}{)}^3}{6.67\times 10^{-11}\times (3.15\times 10^7{)}^2}=\frac{1.33\times 10^{35}}{6.62\times 10^4}$$

About 2.0 × 10³⁰ kg

$\frac{r_s}{r_e}={\left(\frac{T_s}{T_e}\right)}^{2/3}=(29.5{)}^{2/3}=9.55$, so $r_s=9.55\times 1.50\times 10^8$ km $=1.43\times 10^9$ km.

1.43 × 10¹² m

$\frac{W_h}{W}=\frac{R^2}{(R+R/2{)}^2}=\frac{4}{9}$, so $W_h=\frac{4}{9}\times 63$.

28 N

$g(d)=g\left(1-\frac{d}{R}\right)$ with $d=\frac{R}{2}$: $g(d)=\frac{g}{2}$, so the weight is $\frac{250}{2}$.

125 N

Energy conservation: $\frac{1}{2}m{v}^2-\frac{GMm}{R}=-\frac{GMm}{r}\Rightarrow \frac{1}{r}=\frac{1}{R}-\frac{v^2}{2GM}$.

$GM=4.0\times 10^{14}$; $\frac{v^2}{2GM}=\frac{2.5\times 10^7}{8.0\times 10^{14}}=3.12\times 10^{-8}$; $\frac{1}{R}=1.5625\times 10^{-7}$.

$\frac{1}{r}=1.25\times 10^{-7}\Rightarrow r=8.0\times 10^6$ m, i.e. $1.6\times 10^6$ m above the surface.

8.0 × 10⁶ m from the earth's centre (1.6 × 10⁶ m above the surface)

$\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_e^2=\frac{1}{2}m{v}_{\infty }^2$ (since $\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$). With $v=3v_e$: $v_{\infty }=\sqrt{9v_e^2-v_e^2}=\sqrt{8}{v}_e=2.83\times 11.2$.

31.7 km s⁻¹

$r=6.4\times 10^6+0.4\times 10^6=6.8\times 10^6$ m. Orbital energy $E=-\frac{GMm}{2r}$; to escape, E must be raised to 0:

$\Delta E=\frac{GMm}{2r}=\frac{6.67\times 10^{-11}\times 6.0\times 10^{24}\times 200}{2\times 6.8\times 10^6}=5.9\times 10^9$ J.

5.9 × 10⁹ J

Initial separation $d=10^{12}$ m; at collision the centres are $2R=2\times 10^7$ m apart. Energy conservation (each star gets speed v):

$-\frac{G{M}^2}{d}=2\cdot \frac{1}{2}M{v}^2-\frac{G{M}^2}{2R}\Rightarrow v^2=GM\left(\frac{1}{2R}-\frac{1}{d}\right)$

$v^2=6.67\times 10^{-11}\times 2\times 10^{30}\times (5\times 10^{-8}-10^{-12})=6.67\times 10^{12}\Rightarrow v=2.6\times 10^6{\text{m s}}^{-1}$.

2.6 × 10⁶ m s⁻¹

At the midpoint the two pulls are equal and opposite, so the net gravitational field (force per unit mass) is zero.

Potential $V=-\frac{GM}{r}-\frac{GM}{r}=-\frac{2\times 6.67\times 10^{-11}\times 100}{0.5}=-2.7\times 10^{-8}$ J kg⁻¹.

An object there is in equilibrium, but it is unstable: if displaced towards one sphere, that sphere's pull becomes stronger and draws it further away from the midpoint.

Force 0; potential −2.7 × 10⁻⁸ J kg⁻¹; equilibrium, but unstable