Every point mass in the universe attracts every other point mass with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between them — acting along the line joining the two masses.
where $G$ = universal gravitational constant $= 6.674 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ — an extremely small value, which is why gravitational attraction is only noticeable when at least one of the masses involved is astronomically large (like a planet).
Gravitational force is always attractive (never repulsive), acts without needing any medium (works through vacuum), and follows the principle of superposition — the net gravitational force on a mass due to several other masses is simply the vector sum of the individual forces from each.
Calculate the gravitational force of attraction between two 10 kg masses placed 2 m apart.
$F = \dfrac{Gm_1m_2}{r^2} = \dfrac{(6.674\times10^{-11})(10)(10)}{2^2} = \dfrac{6.674\times10^{-9}}{4} \approx 1.67\times10^{-9}\ \text{N}$
This extremely tiny force explains why we don't notice gravitational attraction between everyday objects.
Three point masses, each of 4 kg, are placed at the corners of an equilateral triangle of side 2 m. Calculate the magnitude of the net gravitational force on one of the masses due to the other two.
Force between any pair: $F = \dfrac{Gm^2}{r^2} = \dfrac{(6.674\times10^{-11})(16)}{4} \approx 2.67\times10^{-10}\ \text{N}$
Since the triangle is equilateral, the two forces on any one mass (from the other two) act at 60° to each other. Using the resultant formula: $F_{net} = \sqrt{F^2+F^2+2F^2\cos60°} = F\sqrt{2+2(0.5)} = F\sqrt{3} \approx 2.67\times10^{-10} \times 1.732 \approx 4.62\times10^{-10}\ \text{N}$
Every planet moves around the Sun in an elliptical orbit, with the Sun located at one of the two foci of the ellipse.
The line joining a planet to the Sun sweeps out equal areas in equal intervals of time — this means a planet moves faster when closer to the Sun (at perihelion) and slower when farther away (at aphelion), a direct consequence of the conservation of angular momentum.
The square of a planet's orbital time period is directly proportional to the cube of the semi-major axis of its orbit: $T^2 \propto a^3$.
Explain, using Kepler's second law, why a planet moves fastest when it is closest to the Sun.
Kepler's second law is a direct consequence of conservation of angular momentum, $L = mvr$ (approximately, for near-circular motion), which stays constant since gravity acts along the line joining planet and Sun (producing no torque about the Sun). When the planet is closer to the Sun (smaller $r$), $v$ must increase to keep $L = mvr$ constant — so the planet moves fastest at its closest approach (perihelion) and slowest at its farthest point (aphelion).
Two planets, X and Y, orbit a star with orbital periods of 1 year and 8 years respectively. Using Kepler's third law, compare their orbital radii.
Since $T^2 \propto a^3$: $(T_Y/T_X)^2 = (a_Y/a_X)^3$
$(8/1)^2 = (a_Y/a_X)^3 \Rightarrow 64 = (a_Y/a_X)^3 \Rightarrow a_Y/a_X = 64^{1/3} = 4$
Planet Y's orbital radius is 4 times that of Planet X.
Derived by equating gravitational force to $mg$: $g = \dfrac{GM}{R^2}$, where $M$ and $R$ are Earth's mass and radius. This gives $g \approx 9.8\ \text{m/s}^2$ at the surface.
Gravity decreases with increasing height above the surface.
Gravity also decreases with depth, reaching zero at Earth's centre ($d=R$) — this differs from the height variation, where g would only reach zero at infinite height.
Due to Earth's rotation, effective gravity is slightly reduced by a term depending on latitude, being minimum at the equator and maximum at the poles. Additionally, since Earth is slightly oblate (flattened at the poles, bulging at the equator), the equatorial radius is slightly larger than the polar radius — since $g \propto 1/R^2$, this also makes $g$ marginally smaller at the equator than at the poles.
Calculate the percentage decrease in the value of g at a height equal to 1% of Earth's radius above the surface.
Using $g_h \approx g(1 - 2h/R)$, with $h = 0.01R$: $g_h \approx g(1 - 2(0.01)) = g(1-0.02) = 0.98g$
Percentage decrease $\approx 2\%$
At what depth below Earth's surface will the acceleration due to gravity reduce to exactly half its surface value?
Using $g_d = g(1 - d/R)$, set $g_d = g/2$:
$\dfrac{g}{2} = g\left(1 - \dfrac{d}{R}\right) \Rightarrow \dfrac{1}{2} = 1 - \dfrac{d}{R} \Rightarrow \dfrac{d}{R} = \dfrac{1}{2} \Rightarrow d = \dfrac{R}{2}$
So gravity reduces to half its surface value at a depth equal to half of Earth's radius.
Unlike the simplified $mgh$ formula (valid only for small heights near Earth's surface), the general gravitational potential energy between two masses, valid at any separation, is:
$$U = -\dfrac{GMm}{r}$$The negative sign reflects the convention that gravitational PE is taken as zero at infinite separation, and becomes increasingly negative as the masses come closer together (since gravity does positive work pulling them together, this energy is "released," making the system's potential energy more negative).
Calculate the gravitational potential energy of a 500 kg satellite at a distance of $7 \times 10^6$ m from Earth's centre. (M_Earth $= 6 \times 10^{24}$ kg, G $= 6.674\times10^{-11}$ N m²/kg²)
$U = -\dfrac{GMm}{r} = -\dfrac{(6.674\times10^{-11})(6\times10^{24})(500)}{7\times10^6}$
$U \approx -\dfrac{2.002\times10^{17}}{7\times10^6} \approx -2.86\times10^{10}\ \text{J}$
The binding energy of a satellite (or any orbiting body) is the minimum energy required to completely free it from the gravitational pull of the body it orbits, taking it to infinity with zero final kinetic energy — numerically equal in magnitude, but opposite in sign, to its total mechanical energy: $E_{binding} = -E_{total} = \dfrac{GMm}{2r}$ (for a circular orbit).
A satellite of mass 200 kg orbits Earth at a radius where GM/r = 5×10⁷ (SI units). Calculate its total mechanical energy and binding energy.
Total mechanical energy for a circular orbit: $E_{total} = -\dfrac{GMm}{2r} = -\dfrac{(200)(5\times10^7)}{2} = -5\times10^9\ \text{J}$
Binding energy $= -E_{total} = 5\times10^9\ \text{J}$ — this is the energy that would need to be supplied to free the satellite completely from Earth's gravity.
Escape velocity is the minimum speed an object needs (at the surface, without further propulsion) to completely escape a planet's gravitational influence, reaching infinity with exactly zero remaining kinetic energy.
For Earth, $v_e \approx 11.2\ \text{km/s}$.
Escape velocity does not depend on the mass of the escaping object — both a small pebble and a massive rocket need exactly the same escape velocity to leave a given planet, since the object's own mass cancels out algebraically when deriving the formula from energy conservation.
Derive the escape velocity formula using the principle of conservation of energy.
For an object to just barely escape (reaching infinity with zero final velocity/KE), total initial energy must equal total final energy (which is zero, since both KE and PE are zero at infinity):
$\tfrac{1}{2}mv_e^2 + \left(-\dfrac{GMm}{R}\right) = 0$
$\tfrac{1}{2}mv_e^2 = \dfrac{GMm}{R} \Rightarrow v_e = \sqrt{\dfrac{2GM}{R}}$
The Moon's mass is about 1/81 that of Earth, and its radius is about 1/4 that of Earth's. Estimate the ratio of the Moon's escape velocity to Earth's.
$\dfrac{v_{e,Moon}}{v_{e,Earth}} = \sqrt{\dfrac{M_{Moon}/R_{Moon}}{M_{Earth}/R_{Earth}}} = \sqrt{\dfrac{M_{Moon}}{M_{Earth}} \times \dfrac{R_{Earth}}{R_{Moon}}} = \sqrt{\dfrac{1}{81} \times 4} = \sqrt{\dfrac{4}{81}} = \dfrac{2}{9} \approx 0.22$
The Moon's escape velocity is only about 22% of Earth's (roughly 2.4 km/s compared to Earth's 11.2 km/s) — explaining why the Moon cannot retain a substantial atmosphere, since gas molecules more easily reach and exceed this much lower escape velocity.
For a satellite in a stable circular orbit, gravity itself provides the necessary centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv_o^2}{r}$, giving the required orbital velocity:
$$v_o = \sqrt{\dfrac{GM}{r}}$$Notice this matches the structure of Kepler's third law ($T^2 \propto r^3$), confirming that Kepler's empirically-derived law is a direct consequence of Newton's law of gravitation.
For the same radius, $v_e = \sqrt{2}\,v_o$ — escape velocity is always $\sqrt{2}$ times the orbital velocity at that same distance, since escaping requires exactly twice the kinetic energy needed just to maintain a circular orbit at that radius.
Calculate the orbital velocity of a satellite orbiting Earth at a height where the orbital radius is $7 \times 10^6$ m. (M_Earth $= 6\times10^{24}$ kg, G $= 6.674\times10^{-11}$ N m²/kg²)
$v_o = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{(6.674\times10^{-11})(6\times10^{24})}{7\times10^6}} = \sqrt{\dfrac{4.00\times10^{14}}{7\times10^6}} = \sqrt{5.72\times10^7} \approx 7563\ \text{m/s} \approx 7.56\ \text{km/s}$
Two satellites orbit Earth at radii r and 4r respectively. Compare their orbital velocities.
Since $v_o \propto 1/\sqrt{r}$: $\dfrac{v_{o,1}}{v_{o,2}} = \sqrt{\dfrac{4r}{r}} = \sqrt{4} = 2$
The satellite orbiting at radius $r$ has twice the orbital velocity of the satellite orbiting at radius $4r$.
A geostationary satellite orbits Earth in the equatorial plane, in the same direction as Earth's rotation, with a time period exactly equal to Earth's rotational period (about 24 hours) — this makes it appear stationary relative to a fixed point on Earth's surface, which is why it's used extensively for communication and broadcasting satellites, allowing fixed ground antennas to stay pointed at one location in the sky.
A polar satellite orbits at a much lower altitude, passing over (or near) both poles in each orbit, completing many orbits per day. Since Earth rotates beneath it, a polar satellite eventually covers the entire surface of the Earth over successive orbits — ideal for weather monitoring, remote sensing, and surveillance applications requiring full-Earth coverage.
Explain why a geostationary satellite must orbit specifically in the equatorial plane, and not, for example, in a polar orbital plane.
For a satellite to appear stationary relative to a point on Earth's surface, both its orbital period AND the geometry of its orbit must exactly match Earth's own rotation — this is only possible if the satellite's orbit lies in the same plane as Earth's equator (rotating axis perpendicular to this plane). An orbit tilted at any other angle (like a polar orbit) would cause the satellite's ground track to shift north-south over time, never remaining fixed above a single point, even if its time period happened to match 24 hours.
Weightlessness is the sensation of having no apparent weight, experienced when an object (or person) is in a state of free fall — this occurs, for example, for astronauts inside an orbiting satellite or space station, since both the satellite and its occupants are continuously falling towards Earth under gravity at exactly the same rate, and there's no relative normal force between the person and their surroundings.
Astronauts in orbit are not weightless because "there's no gravity in space" — gravity at typical low-Earth orbital altitudes is only slightly weaker than at the surface (often 85-90% of surface gravity). The true cause of weightlessness is that the astronaut and their spacecraft are both in continuous free fall together (following the same curved orbital path), so there's no relative force pressing the astronaut against any surface — this is exactly the same physical situation as feeling weightless during the brief free-fall of a fast-descending lift.
Explain why an astronaut inside the International Space Station (orbiting at a height where gravity is still fairly strong) experiences apparent weightlessness.
The space station and everything inside it (including the astronaut) are in continuous orbital free fall around Earth — both the station and the astronaut accelerate towards Earth's centre at exactly the same rate (determined only by their shared distance from Earth, since gravitational acceleration is independent of mass). Since there's no relative acceleration between the astronaut and the station's interior surfaces, there's no normal (contact) force between them, and the astronaut experiences apparent weightlessness — not because gravity is absent, but because they and their surroundings are falling together at the same rate.