Gravitation and Satellites

Newton's Law of Universal Gravitation

Every Point Mass attracts every other point mass by a force acting along the Line of Action of their centers.

$$F = G \frac{m_1 m_2}{r^2}$$

The force is proportional to the product of masses and inversely proportional to the square of the distance between them.

$F$=Gravitational force(N)

$G$=Universal Gravitational Constant(Nm²/kg²)

$m_1, m_2$=Masses of the objects(kg)

$r$=Distance between centers(m)

$r \to \infty$

→

Force approaches zero

Double distance

→

Force becomes one-fourth

Attractive Force: Gravity is always an attractive force, never repulsive.

Inverse Square Law: The force drops off rapidly as distance increases, following $1/r^2$.

G Value: $G = 6.674 \times 10^{-11}$ Nm²/kg² — extremely small, so gravitational force is only noticeable for astronomical masses.

Proportionality Shorthand: $F \propto m_1 m_2 / r^2$ — useful for ratio problems where absolute values cancel out.

Gravitational Field Strength (g)

The gravitational field strength at a point is the gravitational force exerted per unit mass on a test mass placed at that point.

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

Determines the acceleration due to gravity at a distance r from a central mass M.

$g$=Field strength(N/kg or m/s²)

$M$=Mass of the source planet/star(kg)

Radial Field: Field lines point towards the center of mass in a spherical distribution.

Superposition Principle: Total field is the vector sum of individual fields from multiple masses.

Surface Gravity: At the surface, $r = R$ (planet radius), so $g = GM/R^2$.

Proportionality: $g \propto M/R^2$ — for comparing planets: $g_1/g_2 = (M_1/M_2)(R_2/R_1)^2$.

Satellite Motion

A satellite maintains its orbit when the gravitational force provides the necessary Centripetal Force.

$$v = \sqrt{\frac{GM}{r}}$$

Calculates the stable orbital speed required for a circular orbit.

$v$=Orbital velocity(m/s)

$r$=Orbital Radius ($R_{planet} + Altitude$)(m)

Mass Independence: The speed required for orbit does not depend on the satellite's mass.

Free Fall: Orbiting is effectively a continuous state of free fall 'around' the planet.

Proportionality: $v \propto 1/\sqrt{r}$ — higher orbit means slower speed.

Orbital Radius: $r = R_{planet} + h$ — the full distance from the planet's center, not just the altitude above the surface.

Kepler's Third Law

The square of the Orbital Period (T) is directly proportional to the cube of the average orbital radius (r).

$$T^2 = \left( \frac{4\pi^2}{GM} \right) r^3$$

Relates the orbital period to the orbital radius for any object orbiting a central mass M.

$T$=Orbital period(s)

$r$=Orbital radius (center-to-center)(m)

$M$=Mass of central body(kg)

Proportionality Constant: The term in brackets remains constant for all satellites orbiting the same central body.

Outer Orbits: Satellites further away take significantly longer to complete one revolution.

Ratio Form: For comparing two orbits, $T_1/T_2 = (r_1/r_2)^{3/2}$ — absolute values of $G$ and $M$ cancel out.

Satellite Types

Satellites are classified by their orbital path and synchronization with Earth's rotation.

Geostationary Orbit: High altitude (≈ 36,000 km) orbit in the Equatorial Plane with a 24-hour period.

Low Earth Orbit (LEO): Lower altitude, faster speed, used for polar mapping or high-speed data.

Synchronous Rotation: Geostationary satellites stay fixed over one spot due to matching Earth's rotation.

Coverage: Three geostationary satellites, each covering 120° of longitude, can cover the entire populated Earth.

The orbital radius of a Geostationary Orbit is derived by equating orbital speed with the speed needed to complete one revolution in exactly 24 hours.

$$r = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$$

Calculates the unique orbital radius where the satellite's period equals Earth's rotation period.

$r$=Geostationary orbital radius(m)

$T$=Period (24 hours = 86,400 s for geostationary)(s)

Derivation: Set $v = \sqrt{GM/r}$ equal to $v = 2\pi r / T$, square both sides, solve for $r$.

Height Above Surface: $h = r - R_E \approx 42{,}300 - 6{,}400 = 35{,}900$ km ≈ 36,000 km.

Unique Radius: There is only one possible radius for a geostationary orbit — it is fixed by Earth's mass and rotation period.

Real and Apparent Weight

The real weight of an object is the gravitational pull $w = mg$. The apparent weight is the reading on a scale — what the object 'feels' — and can differ from the real weight in an accelerating system.

$$T = m(g \pm a)$$

The scale reading (tension T) depends on whether the system accelerates upward or downward.

$T$=Apparent weight (scale reading / tension)(N)

$a$=Acceleration of the system (lift)(m/s²)

$a = 0$ (at rest or constant velocity)

→

$T = mg$ — apparent weight equals real weight

Accelerating upward ($a$ up)

→

$T = m(g + a)$ — feels heavier

Accelerating downward ($a$ down)

→

$T = m(g - a)$ — feels lighter

$a = g$ (free fall)

→

$T = 0$ — complete Weightlessness

Lift at Rest: $T = w = mg$. Scale shows true weight.

Lift Accelerating Up: $T = m(g + a) > mg$. Person feels heavier.

Lift Accelerating Down: $T = m(g - a) < mg$. Person feels lighter.

Free Fall: $a = g \Rightarrow T = 0$. Complete weightlessness.

Weightlessness in Satellites

An orbiting satellite is in continuous Free Fall toward the Earth. Everything inside it — astronauts, tools, water — falls at the same rate, so there is no relative force between objects. This creates a gravity-free system where objects appear weightless.

Not Zero Gravity: Gravity still acts on the satellite (it's what keeps it in orbit). The 'weightlessness' is because everything falls together.

Projectile Analogy: A satellite is a projectile moving fast enough that the Earth's surface curves away as fast as it falls.

Gravity-Free System: No force is needed to hold an object in place inside the spacecraft — it naturally follows the same orbit.

Artificial Gravity

To simulate gravity in a weightless spacecraft, the ship is set into rotation around its own axis. The outer wall pushes inward on the astronaut (centripetal force), mimicking the feel of a gravitational pull toward the 'floor'.

$$f = \frac{1}{2\pi}\sqrt{\frac{g}{R}}$$

The rotation frequency needed so that centripetal acceleration at the rim equals $g$.

$f$=Rotation frequency(Hz (rev/s))

$R$=Radius of the spacecraft (distance from axis to floor)(m)

$g$=Desired artificial gravitational acceleration(m/s²)

Derivation: Set centripetal acceleration $a_c = R(2\pi f)^2 = g$, solve for $f$.

Larger Radius = Slower Spin: A bigger spacecraft needs to rotate less frequently to achieve the same artificial $g$.

Limiting Case: If $R \to \infty$, $f \to 0$ — an infinitely large ring barely needs to rotate.