Gravitational Fields — Diagnostic Tests

Unit Tests

UT-1: Gravitational Field Strength Inside a Uniform Sphere

Question:

A planet of mass $M = 6.0 \times 10^{24}\,\text{kg}$ and radius $R = 6.4 \times 10^6\,\text{m}$ is modelled as a uniform sphere. Take $G = 6.67 \times 10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2}$.

(a) Derive the expression for the gravitational field strength at a distance $r$ from the centre where $r \lt R$.

(b) Calculate the field strength at $r = R/2$ and at $r = R$.

(c) A tunnel is drilled through the planet along a diameter. A small object is released from rest at the surface. Show that the object undergoes SHM and calculate the period.

Solution:

(a) For a uniform sphere of density $\rho$, the mass enclosed within radius $r$ is:

$M_{\text{enc}} = \frac{4}{3}\pi r^3 \rho = M\frac{r^3}{R^3}$

By the shell theorem, only the enclosed mass contributes to the gravitational field at radius $r$:

$g(r) = \frac{GM_{\text{enc}}}{r^2} = \frac{GM}{R^3}r$

The field strength increases linearly with $r$ inside the sphere.

(b) At $r = R/2$:

$g = \frac{GM}{R^3} \times \frac{R}{2} = \frac{GM}{2R^2} = \frac{g_{\text{surface}}}{2}$

Surface gravity: $g_s = \frac{GM}{R^2} = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{(6.4 \times 10^6)^2} = \frac{4.002 \times 10^{14}}{4.096 \times 10^{13}} = 9.77\,\text{m}\,\text{s}^{-2}$

At $r = R/2$: $g = 9.77/2 = 4.89\,\text{m}\,\text{s}^{-2}$

At $r = R$: $g = 9.77\,\text{m}\,\text{s}^{-2}$

(c) Inside the tunnel, the gravitational force on the object at distance $x$ from the centre (along the tunnel) is:

$F = -\frac{GMm}{R^3}x$

$a = -\frac{GM}{R^3}x$

This is SHM with $\omega^2 = GM/R^3$.

Period: $T = 2\pi\sqrt{\frac{R^3}{GM}}$

$T = 2\pi\sqrt{\frac{(6.4 \times 10^6)^3}{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}} = 2\pi\sqrt{\frac{2.621 \times 10^{20}}{4.002 \times 10^{14}}} = 2\pi\sqrt{6.551 \times 10^5}$

$= 2\pi \times 809.4 = 5087\,\text{s} \approx 84.8\,\text{minutes}$

Note: this period is independent of the tunnel direction (it only depends on $R^3/GM$, not on the chord chosen). It equals the period of a low-altitude circular orbit.

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UT-2: Gravitational Potential is Always Negative

Question:

(a) Explain why gravitational potential is defined as negative and why it approaches zero at infinity.

(b) Calculate the gravitational potential at the surface of the Earth and at a height of $3R_E$ above the surface.

(c) Calculate the minimum energy required to move a satellite of mass $500\,\text{kg}$ from a circular orbit of radius $r_1 = 2R_E$ to one of radius $r_2 = 5R_E$.

Take $M_E = 6.0 \times 10^{24}\,\text{kg}$, $R_E = 6.4 \times 10^6\,\text{m}$, $G = 6.67 \times 10^{-11}\,\text{N}\,\text{m}^2\,\text{kg}^{-2}$.

Solution:

(a) Gravitational potential at distance $r$ from mass $M$ is:

$V = -\frac{GM}{r}$

The zero of potential is defined at infinity (where the gravitational field is zero). Since work must be done against gravity to move a mass from distance $r$ to infinity, the potential at $r$ must be lower than at infinity. Since $V(\infty) = 0$ and $V(r) \lt V(\infty)$, the potential is negative at all finite distances.

The potential is a scalar quantity (unlike field strength, which is a vector). It represents the gravitational PE per unit mass. A negative potential means a mass at that position has negative PE relative to infinity -- it is "bound" and would need energy input to escape.

(b) At the surface ($r = R_E$):

$V_s = -\frac{GM_E}{R_E} = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6.4 \times 10^6} = -\frac{4.002 \times 10^{14}}{6.4 \times 10^6} = -6.25 \times 10^7\,\text{J}\,\text{kg}^{-1}$

At height $3R_E$ ($r = 4R_E$):

$V = -\frac{GM_E}{4R_E} = -\frac{6.25 \times 10^7}{4} = -1.56 \times 10^7\,\text{J}\,\text{kg}^{-1}$

(c) Total energy in orbit of radius $r$: $E = -\frac{GMm}{2r}$

At $r_1 = 2R_E$:

$E_1 = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 500}{2 \times 2 \times 6.4 \times 10^6} = -\frac{2.001 \times 10^{17}}{2.56 \times 10^7} = -7.82 \times 10^9\,\text{J}$

At $r_2 = 5R_E$:

$E_2 = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 500}{2 \times 5 \times 6.4 \times 10^6} = -\frac{2.001 \times 10^{17}}{6.4 \times 10^7} = -3.13 \times 10^9\,\text{J}$

Energy required: $\Delta E = E_2 - E_1 = -3.13 \times 10^9 - (-7.82 \times 10^9) = 4.69 \times 10^9\,\text{J}$

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UT-3: Escape Velocity Derivation and Application

Question:

(a) Derive the expression for escape velocity from the surface of a planet of mass $M$ and radius $R$.

(b) Calculate the escape velocity from the surface of the Moon ($M_M = 7.35 \times 10^{22}\,\text{kg}$, $R_M = 1.74 \times 10^6\,\text{m}$).

(c) A student argues that "a projectile launched at escape velocity will escape to infinity and then stop." Explain why this is misleading.

Solution:

(a) To escape, the projectile must reach infinity with zero kinetic energy (minimum energy condition). By conservation of energy:

$\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 + 0$

$\frac{1}{2}mv_e^2 = \frac{GMm}{R}$

$v_e = \sqrt{\frac{2GM}{R}}$

This is independent of the mass of the projectile.

(b) $v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^6}}$

$= \sqrt{\frac{9.805 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.635 \times 10^6} = 2374\,\text{m}\,\text{s}^{-1}$

$v_e \approx 2.37\,\text{km}\,\text{s}^{-1}$

(c) The student's statement is misleading for two reasons:

1. "At infinity" is an idealisation. In practice, escape velocity means the object reaches a distance where the gravitational influence of the planet is negligible compared to other bodies. It never truly "reaches infinity."

2. More precisely, escape velocity is the speed at which the total energy (KE + PE) is exactly zero. The object asymptotically approaches zero speed as $r \to \infty$, but never actually stops at any finite distance. At any finite distance, the object still has some residual KE.

The escape velocity is a threshold: above this speed, the object is unbound (hyperbolic trajectory); below it, the object is bound (elliptical trajectory); at exactly escape velocity, the trajectory is parabolic.

Integration Tests

IT-1: Geostationary Orbit Derivation (with Circular Motion and Energy)

Question:

(a) Derive the condition for a geostationary orbit, showing that the orbital radius and period are uniquely determined by the planet's mass.

(b) Calculate the orbital radius, altitude, and orbital speed of a geostationary satellite around the Earth.

(c) Calculate the total energy of the satellite in this orbit, and the energy required to place it there from the Earth's surface.

Solution:

(a) For a geostationary orbit, the satellite must:

• Orbit in the equatorial plane
• Have the same angular velocity as the Earth's rotation
• Be at a height where the orbital period equals one sidereal day

Equating gravitational force to centripetal force:

$\frac{GMm}{r^2} = m\omega^2 r$

$r^3 = \frac{GM}{\omega^2}$

The orbital period is $T = 2\pi/\omega$, so $\omega = 2\pi/T$ where $T = 86164\,\text{s}$ (one sidereal day).

$r^3 = \frac{GMT^2}{4\pi^2}$

This uniquely determines $r$ given $M$ and $T$.

(b) $r^3 = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times (86164)^2}{4\pi^2}$

$= \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 7.424 \times 10^9}{39.48}$

$= \frac{2.972 \times 10^{24}}{39.48} = 7.528 \times 10^{22}$

$r = (7.528 \times 10^{22})^{1/3} = 4.224 \times 10^7\,\text{m}$

Altitude: $h = r - R_E = 4.224 \times 10^7 - 6.4 \times 10^6 = 3.584 \times 10^7\,\text{m} = 35840\,\text{km}$

Orbital speed: $v = \frac{2\pi r}{T} = \frac{2\pi \times 4.224 \times 10^7}{86164} = 3076\,\text{m}\,\text{s}^{-1}$

(c) Total energy: $E = -\frac{GMm}{2r}$

For $m = 1000\,\text{kg}$ (typical communications satellite):

$E = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1000}{2 \times 4.224 \times 10^7} = -\frac{4.002 \times 10^{17}}{8.448 \times 10^7} = -4.74 \times 10^9\,\text{J}$

Energy from the surface (at rest):

On surface: $E_0 = -\frac{GMm}{R_E} = -\frac{4.002 \times 10^{17}}{6.4 \times 10^6} = -6.25 \times 10^{10}\,\text{J}$

Energy required: $\Delta E = E - E_0 = -4.74 \times 10^9 - (-6.25 \times 10^{10}) = 5.78 \times 10^{10}\,\text{J}$

Note: this is the theoretical minimum. In practice, much more energy is needed due to atmospheric drag, the need to change from the Earth's rotational velocity, and orbit-raising manoeuvres.

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IT-2: Binary Star System (with Circular Motion and Energy)

Question:

Two stars of masses $m_1 = 3.0 \times 10^{30}\,\text{kg}$ and $m_2 = 1.0 \times 10^{30}\,\text{kg}$ orbit their common centre of mass in circular orbits. The separation between them is $d = 2.0 \times 10^{11}\,\text{m}$.

(a) Calculate the orbital radii of each star about the centre of mass.

(b) Calculate the orbital period of the system.

(c) Calculate the total energy of the system and show that it equals $-\frac{Gm_1 m_2}{2d}$.

Solution:

(a) Centre of mass: $m_1 r_1 = m_2 r_2$ and $r_1 + r_2 = d$

$r_1 = \frac{m_2 d}{m_1 + m_2} = \frac{1.0 \times 10^{30} \times 2.0 \times 10^{11}}{4.0 \times 10^{30}} = 5.0 \times 10^{10}\,\text{m}$

$r_2 = d - r_1 = 1.5 \times 10^{11}\,\text{m}$

(b) The gravitational force provides the centripetal force for both stars. Equating for star 1:

$\frac{Gm_1 m_2}{d^2} = m_1 \omega^2 r_1$

$\omega^2 = \frac{Gm_2}{d^2 r_1} = \frac{Gm_2(m_1 + m_2)}{d^2 \times m_2 d} = \frac{G(m_1 + m_2)}{d^3}$

$\omega = \sqrt{\frac{G(m_1 + m_2)}{d^3}} = \sqrt{\frac{6.67 \times 10^{-11} \times 4.0 \times 10^{30}}{8.0 \times 10^{33}}}$

$= \sqrt{\frac{2.668 \times 10^{20}}{8.0 \times 10^{33}}} = \sqrt{3.335 \times 10^{-14}} = 1.826 \times 10^{-7}\,\text{rad}\,\text{s}^{-1}$

Period: $T = 2\pi/\omega = 2\pi/(1.826 \times 10^{-7}) = 3.441 \times 10^7\,\text{s} \approx 398\,\text{days}$

(c) Total KE:

$KE = \frac{1}{2}m_1(\omega r_1)^2 + \frac{1}{2}m_2(\omega r_2)^2 = \frac{1}{2}\omega^2(m_1 r_1^2 + m_2 r_2^2)$

Total PE: $PE = -\frac{Gm_1 m_2}{d}$

Using the virial theorem for gravitational systems ($KE = -\frac{1}{2}PE$):

$E = KE + PE = -\frac{1}{2}PE + PE = \frac{1}{2}PE = -\frac{Gm_1 m_2}{2d}$

Substituting: $E = -\frac{6.67 \times 10^{-11} \times 3.0 \times 10^{30} \times 1.0 \times 10^{30}}{2 \times 2.0 \times 10^{11}}$

$= -\frac{2.001 \times 10^{50}}{4.0 \times 10^{11}} = -5.00 \times 10^{38}\,\text{J}$

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IT-3: Gravitational Potential and Field Between Two Planets (with Energy)

Question:

Two identical planets each of mass $M = 3.0 \times 10^{24}\,\text{kg}$ and radius $R = 4.0 \times 10^6\,\text{m}$ are separated by a distance $d = 5.0 \times 10^7\,\text{m}$ (centre to centre).

(a) Calculate the gravitational potential at the midpoint between the two planets.

(b) Calculate the position (between the planets) where the gravitational field strength is zero.

(c) A space probe of mass $100\,\text{kg}$ is at the point of zero field strength. Calculate the minimum speed it must have to reach the surface of either planet.

Solution:

(a) At the midpoint ($r = d/2 = 2.5 \times 10^7\,\text{m}$ from each planet):

$V = -\frac{GM}{d/2} - \frac{GM}{d/2} = -\frac{4GM}{d}$

$V = -\frac{4 \times 6.67 \times 10^{-11} \times 3.0 \times 10^{24}}{5.0 \times 10^7} = -\frac{8.004 \times 10^{14}}{5.0 \times 10^7} = -1.60 \times 10^7\,\text{J}\,\text{kg}^{-1}$

(b) Let the zero-field point be at distance $x$ from planet 1 (so $d - x$ from planet 2):

$\frac{GM}{x^2} = \frac{GM}{(d - x)^2}$

$x^2 = (d - x)^2$

$x = d - x \Rightarrow x = d/2 = 2.5 \times 10^7\,\text{m}$

This is the midpoint (expected by symmetry since the planets are identical).

(c) At the midpoint, the potential energy of the probe is:

$PE_{\text{mid}} = mV = 100 \times (-1.60 \times 10^7) = -1.60 \times 10^9\,\text{J}$

At the surface of planet 1 (distance $x = d/2 - R = 2.5 \times 10^7 - 4.0 \times 10^6 = 2.1 \times 10^7\,\text{m}$ from the midpoint):

$PE_{\text{surf}} = -\frac{GMm}{R} - \frac{GMm}{d - R}$

$= -\frac{6.67 \times 10^{-11} \times 3.0 \times 10^{24} \times 100}{4.0 \times 10^6} - \frac{6.67 \times 10^{-11} \times 3.0 \times 10^{24} \times 100}{4.6 \times 10^7}$

$= -\frac{2.001 \times 10^{16}}{4.0 \times 10^6} - \frac{2.001 \times 10^{16}}{4.6 \times 10^7}$

$= -5.00 \times 10^9 - 4.35 \times 10^8 = -5.44 \times 10^9\,\text{J}$

The probe needs to go from PE $= -1.60 \times 10^9\,\text{J}$ to PE $= -5.44 \times 10^9\,\text{J}$, a decrease of $3.84 \times 10^9\,\text{J}$.

Since PE decreases (becomes more negative), the probe gains KE. It needs zero initial speed -- gravity will pull it to the surface. The probe at the zero-field point is at a gravitational potential "ridge" (unstable equilibrium). Any perturbation towards either planet will cause it to fall. The minimum speed is zero (it is at an unstable equilibrium point).