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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.
Application Strategy
Worked Examples
Problem Statement
Calculate the gravitational force between a 60 kg student and a 70 kg student sitting 2.0 meters apart.
Newton's Law of Gravitation
Strategic Toolkit
$F = G \frac{m_1 m_2}{r^2}$
Step 1
Identify Variables
Given $m_1 = 60$ kg, $m_2 = 70$ kg, $r = 2$ m, $G = 6.67 \times 10^{-11}$ Nm²/kg².
$m_1=60, m_2=70, r=2$
Step 2
Write the Formula
Newton's law of universal gravitation:
$F = G \frac{m_1 m_2}{r^2}$
Step 3
Substitute Known Values
With $G = 6.67 \times 10^{-11}$, $m_1 = 60$ kg, $m_2 = 70$ kg, $r = 2$ m:
$F = \frac{(6.67 \times 10^{-11}) \times 60 \times 70}{2^2}$
Step 4
Calculate
Evaluate the expression:
$F = \frac{6.67 \times 10^{-11} \times 4200}{4} = 7.0 \times 10^{-8}$ N
Problem Statement
Determine the orbital speed of the ISS at an altitude of 400 km. ($R_{Earth} = 6400$ km, $M_{Earth} = 5.97 \times 10^{24}$ kg)
Orbital VelocityRadius vs Altitude
Strategic Toolkit
$v = \sqrt{\frac{GM}{r}}$
Step 1
Determine Total Radius
Radius $r$ must include altitude and Earth's radius.
$r = 6400 \text{ km} + 400 \text{ km} = 6800 \text{ km} = 6.8 \times 10^6$ m
Step 2
Write the Orbital Speed Formula
For a circular orbit where gravity provides centripetal force:
$v = \sqrt{\frac{GM}{r}}$
Step 3
Substitute Known Values
With $G = 6.67 \times 10^{-11}$, $M = 5.97 \times 10^{24}$ kg, $r = 6.8 \times 10^6$ m:
$v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.8 \times 10^6}}$
Step 4
Calculate
Evaluate the expression:
$v = \sqrt{\frac{3.98 \times 10^{14}}{6.8 \times 10^6}} = \sqrt{5.86 \times 10^7} \approx 7652$ m/s
Problem Statement
Calculate the orbital radius and speed of a geostationary satellite. ($M_{Earth} = 6.0 \times 10^{24}$ kg, $T = 24$ h)
Geostationary OrbitKepler's Third Law
Strategic Toolkit
$r = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$
Step 1
Convert Period to Seconds
The period must be in SI units.
$T = 24 \times 3600 = 86{,}400$ s
Step 2
Write the Geostationary Radius Formula
From Kepler's Third Law, solved for $r$:
$r = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$
Step 3
Substitute Known Values
With $G = 6.67 \times 10^{-11}$, $M = 6.0 \times 10^{24}$ kg, $T = 86{,}400$ s:
$r = \left( \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times (86400)^2}{4\pi^2} \right)^{1/3}$
Step 4
Calculate Radius
Evaluate the expression:
$r = \left( \frac{4.0 \times 10^{14} \times 7.46 \times 10^{9}}{39.48} \right)^{1/3} = 4.23 \times 10^7$ m
Step 5
Calculate Orbital Speed
Use $v = 2\pi r / T$:
$v = \frac{2\pi \times 4.23 \times 10^7}{86400} \approx 3.1$ km/s
Problem Statement
A 70 kg person stands on a scale in a lift accelerating upward at 2 m/s². What does the scale read?
Apparent WeightNewton's Second Law
Strategic Toolkit
T = m(g + a)
Step 1
Identify Variables
Mass $m = 70$ kg, upward acceleration $a = 2$ m/s², $g = 9.8$ m/s².
$m = 70, \; a = 2, \; g = 9.8$
Step 2
Apply Apparent Weight Formula
Lift accelerating upward: $T = m(g + a)$.
$T = 70(9.8 + 2) = 70 \times 11.8 = 826$ N
Step 3
Compare to Real Weight
Real weight $= mg = 70 \times 9.8 = 686$ N. The scale reads 826 N — 140 N more than the real weight.
$\Delta W = ma = 70 \times 2 = 140$ N heavier
Problem Statement
Satellite A orbits at radius $r$ and satellite B at radius $4r$ around the same planet. Find $T_B / T_A$.
Kepler's Third LawProportionality
Strategic Toolkit
$T \propto r^{3/2}$
Step 1
Set Up Ratio
Using $T \propto r^{3/2}$, take the ratio.
$\frac{T_B}{T_A} = \left( \frac{r_B}{r_A} \right)^{3/2} = \left( \frac{4r}{r} \right)^{3/2}$
Step 2
Calculate
Simplify: $4^{3/2} = (\sqrt{4})^3 = 2^3 = 8$.
$T_B / T_A = 8$
Memorize formulas
Universal Gravitation
Inverse square between two masses
Formula
$F = G \frac{m_1 m_2}{r^2}$
Gravitational Field
Strength at distance r
Formula
$g = \frac{GM}{r^2}$
Orbital Speed
Balance of gravity and circular motion
Formula
$v = \sqrt{\frac{GM}{r}}$
Kepler's Law
T-squared over r-cubed
Formula
$\frac{T^2}{r^3} = \frac{4\pi^2}{GM}$
Geostationary Radius
Unique orbit radius for 24-hour period
Formula
$r = \left( \frac{GMT^2}{4\pi^2} \right)^{1/3}$
Apparent Weight
Scale reading in accelerating lift
Formula
$T = m(g \pm a)$
Artificial Gravity Frequency
Rotation rate for simulated gravity
Formula
$f = \frac{1}{2\pi}\sqrt{\frac{g}{R}}$
Orbital Period
Time for one complete orbit
Formula
$T = \frac{2\pi r}{v}$
Review
1
A satellite is moved from an orbit of radius R to an orbit of radius 2R. By what factor does the gravitational force from the planet change?
Ans
It decreases by a factor of 4
Review
Tap to reveal
Apply inverse-square scaling
Gravitational force varies as $1/r^2$, so doubling the orbital radius makes the force one-quarter of its original value.
$$F' = F\left(\frac{R}{2R}\right)^2 = \frac{F}{4}$$
2
A neutron star forms when a star of mass $M$ and radius $R$ collapses to a radius of $R/1000$ while retaining the same mass. The surface gravity of the neutron star compared to the original star is:
Ans
$10^6$ times larger
Review
Tap to reveal
Use $g = GM/r^2$
The mass stays the same, but the radius shrinks to $R/1000$. Since surface gravity is inversely proportional to $r^2$, the new value increases by $(1000)^2$.
$$g' = \frac{GM}{(R/1000)^2} = 10^6\,\frac{GM}{R^2} = 10^6 g$$
3
Which of the following is mandatory for a geostationary orbit?
Ans
Orbiting the equatorial plane with a 24-hour period
Review
Tap to reveal
Match the defining condition for geostationary orbit
A geostationary satellite must remain above one point on Earth, which requires an equatorial orbit with the same period as Earth's rotation.
$$\text{equatorial orbit},\quad T = 24\text{ h}$$
4
A 60 kg astronaut stands on a scale inside a rocket accelerating upward at $4g$. The scale reading is:
Ans
$5mg = 2940$ N
Review
Tap to reveal
Add the rocket acceleration to gravity
The scale reads the normal force. In an upward-accelerating rocket, that force must support both gravity and the upward acceleration.
$$N = m(g+a) = 60(9.8+4\times 9.8)=2940\text{ N}$$
5
A satellite in orbit around Earth is in a state of weightlessness because:
Ans
The satellite and everything inside it are in free fall together
Review
Tap to reveal
Recognize free fall
The satellite and everything inside it accelerate together under gravity, so there is no supporting contact force between objects inside the craft.
$$\text{weightlessness} \Rightarrow \text{same free-fall acceleration for all objects}$$
6
If the radius of a geostationary orbit is $r$, what is the orbital speed in terms of $r$?
Ans
$v = 2\pi r / (24 \times 3600)$
Review
Tap to reveal
Use speed = distance per period
For one circular revolution, the satellite travels a circumference of $2\pi r$ in one 24-hour period.
$$v = \frac{2\pi r}{24\times 3600}$$
7
Two satellites orbit the same planet. Satellite B has an orbital period 27 times that of satellite A. The ratio of their orbital speeds $v_A/v_B$ is:
Ans
3
Review
Tap to reveal
Combine Kepler's law with $v = 2\pi r/T$
Kepler's third law gives $r_B/r_A = 27^{2/3} = 9$. Substituting into the speed ratio then gives the final factor of 3.
$$\frac{v_A}{v_B} = \frac{r_A/T_A}{r_B/T_B} = \frac{r_A}{r_B}\cdot\frac{T_B}{T_A} = \frac{1}{9}\cdot 27 = 3$$
8
A space station of radius $R$ rotates to create artificial gravity equal to $g$. If the radius is quadrupled to $4R$, the required rotation frequency becomes:
Ans
Half the original
Review
Tap to reveal
Use the frequency-radius relation
Artificial gravity comes from centripetal acceleration $a = (2\pi f)^2 R$. For fixed $a = g$, the rotation frequency is proportional to $1/\sqrt{R}$.
$$g = (2\pi f)^2R \Rightarrow f \propto \frac{1}{\sqrt R} \Rightarrow f' = \frac{f}{2}$$
9
Planet P has mass $4M$ and radius $R$ (same as Earth). Planet Q has mass $M$ and radius $R/2$. The ratio $g_P/g_Q$ is:
Ans
1
Review
Tap to reveal
Compare each surface gravity directly
Surface gravity is $GM/r^2$. Planet P has four times the mass at radius $R$, while planet Q has one-quarter the radius but the same mass, and both evaluate to the same value.
$$g_P = \frac{G(4M)}{R^2},\quad g_Q = \frac{GM}{(R/2)^2} = \frac{4GM}{R^2} \Rightarrow \frac{g_P}{g_Q}=1$$
10
A person in a freely falling lift drops a ball from rest. Relative to the person, the ball will:
Ans
Remain stationary (float)
Review
Tap to reveal
Use the shared free-fall motion
In a freely falling lift, the observer and the released ball both accelerate downward at $g$. With no relative acceleration, the ball stays at rest relative to the person.
$$a_{\text{person}} = a_{\text{ball}} = g \Rightarrow a_{\text{relative}} = 0$$
11
The gravitational force between two spheres is $F$. If the mass of each sphere is doubled AND the distance between them is also doubled, the new force is:
Ans
$F$
Review
Tap to reveal
Track both mass and distance changes
Doubling each mass multiplies the numerator by 4, but doubling the separation multiplies the denominator by 4 as well. The factors cancel exactly.
$$F' = \frac{G(2m_1)(2m_2)}{(2r)^2} = \frac{4Gm_1m_2}{4r^2} = F$$
12
A satellite orbits Earth at altitude $h$ above the surface. Its orbital speed is $v = \sqrt{GM/(R+h)}$. If $h \ll R$ (very low orbit), the speed approximates to:
Ans
$\sqrt{gR}$
Review
Tap to reveal
Approximate the orbit radius by $R$
For very low orbit, $h$ is negligible compared with $R$, so $r \approx R$. Substituting $g = GM/R^2$ converts the speed to $\sqrt{gR}$.
$$v = \sqrt{\frac{GM}{R+h}} \approx \sqrt{\frac{GM}{R}} = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$$
Question 1 of 12
0:00
1
A satellite is moved from an orbit of radius R to an orbit of radius 2R. By what factor does the gravitational force from the planet change?