Gravitation

Acceleration due to gravity is the acceleration with which a freely falling body moves towards the centre of the earth. At the surface $g = 9.8\ \text{m/s}^2$.

Variation with altitude. Let the earth have mass $M$ and radius $R$.

At the surface, $mg=\dfrac{GMm}{R^2}$, so

$g=\frac{GM}{R^2}$

At a height $h$ the distance from the centre is $R+h$, so

$g'=\frac{GM}{(R+h)^2}$

Dividing,

$g'=g\left(\frac{R}{R+h}\right)^2=g\left(1+\frac hR\right)^{-2}$

For $h\ll R$, the binomial expansion gives

$g'\approx g\left(1-\frac{2h}{R}\right)$

So $g$ decreases as the height increases.

(a) Escape velocity is the minimum velocity with which a body must be projected from the surface of a planet so that it just escapes from the planet's gravitational field and never returns.

(b) Work done against gravity to take a mass $m$ from the surface (distance $R$ from the centre) to infinity:

$W=\int_R^{\infty}\frac{GMm}{x^2}\,dx=\frac{GMm}{R}$

This must be supplied by the kinetic energy at projection:

$\tfrac12 m v_e^2=\frac{GMm}{R}\;\Rightarrow\; v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\quad(\text{since } GM=gR^2)$

For the earth, $v_e\approx11.2\ \text{km s}^{-1}$.

(c) Orbital radius $r=R+h=6400+20=6420$ km $=6.42\times10^{6}$ m. Using $GM=gR^2$:

$T=2\pi\sqrt{\frac{r^3}{GM}}=2\pi\sqrt{\frac{r^3}{gR^2}}=2\pi\sqrt{\frac{(6.42\times10^{6})^3}{10\times(6.4\times10^{6})^2}}$

$T=2\pi\sqrt{6.46\times10^{5}}=2\pi\times804\approx5.05\times10^{3}\ \text{s}\approx 84\ \text{minutes}$

(d) Yes, it does. Since $v_e=\sqrt{2GM/R}$ and $M$ is unchanged, $v_e\propto R^{-1/2}$. If $R$ becomes $0.98R$, then $v_e'=v_e/\sqrt{0.98}\approx1.01\,v_e$, so the escape velocity increases by about 1%.

They are related by $g=-\dfrac{dV}{dr}$.

The gravitational force of the earth provides the centripetal force for the moon:

$\frac{GMm}{r^2}=m\left(\frac{2\pi}{T}\right)^2 r\;\Rightarrow\; r^3=\frac{GMT^2}{4\pi^2}$

Data: $T=27.3\times24\times3600=2.36\times10^{6}$ s, $M=5.97\times10^{24}$ kg, $G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$.

$GM=6.67\times10^{-11}\times5.97\times10^{24}=3.98\times10^{14}$

$T^2=(2.36\times10^{6})^2=5.56\times10^{12}$

$r^3=\frac{3.98\times10^{14}\times5.56\times10^{12}}{4\pi^2}=\frac{2.21\times10^{27}}{39.48}=5.6\times10^{25}\ \text{m}^3$

$r=\sqrt[3]{5.6\times10^{25}}\approx3.8\times10^{8}\ \text{m}$

The moon is about $3.8\times10^{8}$ m ($3.8\times10^{5}$ km) from the centre of the earth.

A satellite and the person inside it are in free fall around the earth with the same acceleration, so the floor exerts no normal reaction on the person. The apparent weight (the reaction force) is zero, and the person feels weightless.

A person standing on the moon is not in free fall relative to the moon's surface. The surface supports him with a normal reaction equal to the moon's gravitational pull $mg_m$ (where $g_m\approx 1.6\ \text{m s}^{-2}$), so he feels his weight. The moon itself is weightless only with respect to the earth, not with respect to bodies resting on it.

(d) all bodies irrespective of their size

Newton's law $F=\dfrac{Gm_1m_2}{r^2}$ is universal: every pair of masses in the universe attracts each other, whether they are small objects, planets or stars.

The body must be given just enough kinetic energy to reach infinity, where the total energy is zero:

$\tfrac12 mv_e^2=\frac{GMm}{R}\;\Rightarrow\; v_e=\sqrt{\frac{2GM_e}{R_e}}$

$v_e=\sqrt{\frac{2\times6.67\times10^{-11}\times5.98\times10^{24}}{6.4\times10^{6}}}=\sqrt{1.246\times10^{8}}$

$v_e\approx1.12\times10^{4}\ \text{m/s}\;(\approx 11.2\ \text{km/s})$

The escape velocity does not depend on the mass of the body, so the 1000 kg mass makes no difference to the result. (Its kinetic energy needed is $\tfrac12\times1000\times(1.12\times10^4)^2\approx6.2\times10^{10}$ J.)

GPS is a satellite-based navigation system. A network of about 24 or more satellites orbiting the earth continuously broadcasts timed radio signals. A receiver measures the signal travel time from at least four satellites and uses it to find its distance from each, which fixes its position (latitude, longitude, altitude) and the time.

Applications:

• Navigation of cars, ships, aircraft and mobile-phone maps.
• Surveying and mapping of land, and locating positions in remote areas.
• Tracking vehicles and fleets, and search and rescue after disasters.
• Accurate time synchronisation for communication and power networks.
• Weather forecasting, and monitoring of crustal movement and earthquakes.

(a) Gravitational potential at a point is the work done per unit mass in bringing a small mass from infinity to that point. Gravity is attractive, so the field itself does the work and the mass loses energy as it comes in. Since the potential is defined as zero at infinity, it must be negative everywhere nearer: $V=-\dfrac{GM}{r}$.

(b) $R=6.4\times10^{6}\ \text{m}$; the final distance from the centre is $r=R+h=6.4\times10^{6}+1.3\times10^{7}=1.94\times10^{7}\ \text{m}$ $GM=6.67\times10^{-11}\times6.0\times10^{24}=4.0\times10^{14}\ \text{m}^3\text{s}^{-2}$.

(i) $V_1=-\frac{GM}{R}=-6.25\times10^{7}\ \text{J kg}^{-1},\qquad V_2=-\frac{GM}{r}=-2.06\times10^{7}\ \text{J kg}^{-1}$ $\Delta V=V_2-V_1\approx+4.2\times10^{7}\ \text{J kg}^{-1}$

(ii) The kinetic energy at launch equals the gain in potential energy (it is momentarily at rest at the top): $\tfrac12v^2=\Delta V\;\Rightarrow\;v=\sqrt{2\times4.19\times10^{7}}\approx9.2\times10^{3}\ \text{m s}^{-1}$

(c) $v^2=u^2+2as$ holds only for constant acceleration. Over an altitude of $1.3\times10^4$ km, $g$ falls greatly with distance ($g\propto1/r^2$), so the acceleration is not constant.

(d) increase by 2%

$g=\dfrac{GM}{R^2}$, so with $M$ fixed $g\propto R^{-2}$. A 1% decrease in $R$ gives $\dfrac{\Delta g}{g}=-2\,\dfrac{\Delta R}{R}=+2\%$.

Acceleration due to gravity ($g$) is the acceleration a freely falling body gains due to the Earth's gravitational pull. At the surface $g=\dfrac{GM}{R^2}$, about $9.8\ \text{m s}^{-2}$.

Variation with altitude. At height $h$ above the surface, the distance from the centre is $R+h$:

$g_h=\frac{GM}{(R+h)^2}=g\left(\frac{R}{R+h}\right)^2\approx g\left(1-\frac{2h}{R}\right)\quad(h\ll R)$

So $g$ decreases as we go up.

Variation with depth. At depth $d$, only the inner sphere of radius $R-d$ attracts the body (the outer shell exerts no net force). With uniform density $\rho$, $g\propto$ radius, so

$g_d=g\left(1-\frac{d}{R}\right)$

So $g$ decreases linearly with depth and becomes zero at the centre of the Earth. It is maximum at the surface.

Use $GM=gR_E^2=9.81\times(6.4\times10^{6})^2\approx4.0\times10^{14}\ \text{m}^3\text{s}^{-2}$.

(a)(i) $E_p=mV$, where $V$ is the gravitational potential at the point.

(a)(ii) If $m$ is doubled, $E_p$ doubles, while $V$ is unchanged because potential is energy per unit mass and depends only on the field.

(b)(i) $V=-\dfrac{GM}{r}$, so for orbit A:

$r_A=\frac{GM}{|V_A|}=\frac{4.0\times10^{14}}{12.0\times10^{6}}=3.3\times10^{7}\ \text{m}\approx3.3\times10^{4}\ \text{km}$

(b)(ii) For orbit B:

$r_B=\frac{4.0\times10^{14}}{36.0\times10^{6}}=1.11\times10^{7}\ \text{m}\approx1.1\times10^{4}\ \text{km}$

(b)(iii) The centripetal acceleration equals the local gravitational field:

$a=\frac{GM}{r_B^2}=\frac{4.0\times10^{14}}{(1.11\times10^{7})^2}\approx3.2\ \text{m s}^{-2}$

(b)(iv)

$\Delta E_p=m\,\Delta V=330\times\left[(-36.0)-(-12.0)\right]\times10^{6}=-7.9\times10^{9}\ \text{J}$

The potential energy decreases by about $8\times10^{9}\ \text{J}=8\ \text{GJ}$.

(a) Gravitational potential at a point is the work done by an external agent in bringing a unit mass from infinity to that point without acceleration (equivalently, the gravitational potential energy per unit mass). It is $V=-\dfrac{GM}{r}$. SI unit: $\text{J kg}^{-1}$.

(b) Potential is potential energy per unit mass, so

$V=\frac{E_P}{m}\quad\text{or}\quad E_P=mV$

(c)(i) For a point mass, $V=-\dfrac{GM}{r}\propto\dfrac1r$. At the surface ($r=R$), $V_R=-6.30\times10^{7}\ \text{J kg}^{-1}$. Then

$V_{6R}=\frac{-6.30\times10^{7}}{6}=-1.05\times10^{7},\qquad V_{5R}=\frac{-6.30\times10^{7}}{5}=-1.26\times10^{7}\ \text{J kg}^{-1}$

$\Delta V=-1.26\times10^{7}-(-1.05\times10^{7})=-2.1\times10^{6}\ \text{J kg}^{-1}$

$\Delta E_P=m\Delta V=1.30\times(-2.1\times10^{6})=-2.73\times10^{6}\ \text{J}$

The gravitational potential energy decreases by $2.73\times10^{6}$ J.

(ii) The negative sign means the gravitational force is attractive: the potential is zero at infinity (the maximum value) and decreases as one comes closer to the mass, so work must be done against gravity to take a mass from that point to infinity. The stone is bound to the planet.

(a) A gravitational potential of $-1.70\times10^{9}\ \text{J kg}^{-1}$ at A means that $1.70\times10^{9}\ \text{J}$ of work is done by the gravitational field per kilogram when a mass is brought from infinity to A. Equivalently, $1.70\times10^{9}\ \text{J}$ must be supplied per kilogram to take the mass from A to infinity. The negative sign shows that the field is attractive, since the potential at infinity is taken as zero.

(b)

$V_B=-\frac{GM}{r}=-\frac{(6.67\times10^{-11})(1.90\times10^{27})}{1.69\times10^{8}}=-7.50\times10^{8}\ \text{J kg}^{-1}$

(c)

$\Delta U=m(V_B-V_A)=1.6\times10^{3}\left[(-7.50\times10^{8})-(-1.70\times10^{9})\right]$

$\Delta U=1.6\times10^{3}\times9.5\times10^{8}=+1.52\times10^{12}\ \text{J}$

The potential energy increases by $1.52\times10^{12}\ \text{J}$.

Weight $=mg$, so it depends on the effective value of $g$, which is larger at the poles than at the equator for two reasons.

1. The earth is slightly flattened at the poles, so the polar radius is smaller than the equatorial radius. Since $g=\dfrac{GM}{R^2}$, $g$ is larger at the poles.
2. Because of the earth's rotation, a body at the equator needs a centripetal force $m\omega^2R$, which reduces its effective weight. At the poles the radius of the circular path is zero, so there is no such reduction.

Hence a body weighs more at the pole.

Take $R=6400\ \text{km}$, $g=9.8\ \text{m s}^{-2}$. The orbital radius is

$r=R+h=6400+36000=42400\ \text{km}=4.24\times10^{7}\ \text{m}$

Gravity provides the centripetal force, and $GM=gR^2$, so

$T=2\pi\sqrt{\frac{r^3}{GM}}=2\pi\sqrt{\frac{r^3}{gR^2}}$

$GM=9.8\times(6.4\times10^6)^2=4.01\times10^{14}\ \text{m}^3\text{s}^{-2}$

$r^3=(4.24\times10^7)^3=7.62\times10^{22}\ \text{m}^3$

$T=2\pi\sqrt{\frac{7.62\times10^{22}}{4.01\times10^{14}}}=2\pi\sqrt{1.90\times10^{8}}\approx 8.66\times10^{4}\ \text{s}$

$T=\frac{8.66\times10^4}{3600}\approx 24\ \text{hours}$

The satellite's period equals that of the earth's rotation, so it is geostationary.