The total energy of a satellite in a stable circular orbit is:

Negative, equal in magnitude to its kinetic energy

Positive, equal to its kinetic energy

Negative, equal in magnitude to its kinetic energy

Negative, equal in magnitude to its potential energy

A satellite in a low orbit gradually loses energy to atmospheric drag. Its orbital speed:

Increases, because the orbital radius shrinks

Decreases, since it is losing energy

Increases, because the orbital radius shrinks

Which is NOT a requirement for a geostationary satellite?

A mass equal to that of a standard communication satellite

An orbit in the equatorial plane

A mass equal to that of a standard communication satellite

Astronauts float inside the International Space Station because:

They and the station are in free fall together, so the contact force is zero

There is no gravity at that altitude

They and the station are in free fall together, so the contact force is zero

The station spins to cancel gravity

Their mass becomes zero in orbit

Gravitation — Fields, Potential, Orbits and Escape

Gravitation is a small chapter that reliably yields one or two questions, and it is unusually forgiving because almost everything follows from a single inverse-square law plus one energy equation. The two places students lose marks are both about signs and reference points: gravitational potential energy is negative, which feels wrong until you see where the zero is chosen, and the total energy of an orbit is negative too, which is exactly what "bound" means. Get those two straight and orbital mechanics, escape velocity and satellite energetics all become one calculation.

Newton's Law and the Variation of g

Every pair of point masses attracts along the line joining them with F = Gm₁m₂/r², where G = 6.67 × 10⁻¹¹ N·m²/kg² is a universal constant. Applying it to the Earth gives g = GM/R² at the surface, and the whole of the next table is what happens when you move away from that surface.

How g changes with position
Where	Value	Behaviour
At height h above the surface	g_h = g R²/(R+h)² ≈ g(1 − 2h/R) for h ≪ R	Falls as an inverse square
At depth d below the surface	g_d = g(1 − d/R)	Falls linearly, reaching exactly zero at the centre
With latitude	g_λ = g − ω²R cos²λ	Maximum at the poles, minimum at the equator

🎯 Why g falls linearly going down but quadratically going up

Above the surface the whole mass M stays below you and the distance grows, so g ∝ 1/r² — a simple inverse square. Below the surface something different happens: a shell of matter exerts no net gravitational force on anything inside it, so all the mass in the shell above your depth contributes nothing at all. Only the sphere beneath you pulls, and its mass grows as r³ while the inverse square divides by r², leaving g ∝ r. So g rises linearly from zero at the centre to its maximum at the surface, then falls off as an inverse square outside. A common exam trick pairs "at height h" with "at depth h" and asks which gives the smaller g: for small h the answers are g(1 − 2h/R) and g(1 − h/R), so height reduces g twice as fast as depth.

Potential, Potential Energy and the Negative Sign

Four quantities that are easy to confuse
Quantity	Expression	Nature
Field strength	E = GM/r²	Vector, N/kg — same thing as g
Potential	V = −GM/r	Scalar, J/kg
Potential energy of a mass m	U = −GMm/r	Scalar, J
Relation between them	E = −dV/dr	Field is the negative gradient of potential

🎯 Why the potential energy is negative, and what that buys you

The zero of gravitational potential energy is chosen at infinite separation, where the two bodies no longer interact. Bringing a mass from infinity to distance r means gravity does positive work on it, so its potential energy falls below zero — hence U = −GMm/r. The sign is not a convention you must simply accept; it carries information. Negative total energy means bound, positive means free, and zero is exactly the borderline. So you can answer "will this object escape?" by computing the sign of its total energy and nothing else. Note also that the familiar U = mgh is only the near-surface approximation of this: it measures the change in U over a height small compared with R, with the zero moved to the ground for convenience.

Satellites, Orbital Speed and Escape Velocity

A satellite in a circular orbit of radius r has gravity supplying exactly the centripetal force: GMm/r² = mv²/r, which gives v_orbit = √(GM/r). Every other orbital result follows from that one line together with U = −GMm/r.

The satellite energy table — worth deriving once and remembering
Quantity	Value	Note
Orbital speed	v = √(GM/r)	Independent of the satellite's mass
Kinetic energy	KE = +GMm/2r	—
Potential energy	U = −GMm/r	U = −2 × KE
Total energy	E = −GMm/2r	Negative → bound. E = −KE
Time period	T² = 4π²r³/GM	Kepler's third law, T² ∝ r³
Escape velocity	v_e = √(2GM/R) = √(2gR) ≈ 11.2 km/s	v_e = √2 × v_orbit at the same radius

🧠 The √2 relation, and why escape velocity has no direction

Escape requires total energy ≥ 0, so ½mv² = GMm/R gives v_e = √(2GM/R), which is exactly √2 times the orbital speed at that radius. That single relation answers a lot of questions in one step — a satellite in low orbit needs its speed multiplied by about 1.41, not by 2, to escape. Two further points that get examined. Escape velocity is a speed, not a velocity: since gravity is conservative, only the magnitude matters and the direction of launch is irrelevant (ignoring the atmosphere and the ground). And v_e is independent of the mass of the escaping object, so a spacecraft and a pebble need the same 11.2 km/s — which is why the figure is quoted as a property of the Earth rather than of the rocket.

⚠️ A satellite losing energy speeds up

Air drag in a low orbit removes energy, so E = −GMm/2r becomes more negative, so r decreases — and since v = √(GM/r), a smaller r means a larger speed. So friction makes the satellite go faster, which sounds like a contradiction and is not: the orbit shrinks, potential energy falls by more than the kinetic energy rises, and the total still goes down. The relation to hold on to is that kinetic energy rises while total energy falls, and the two are linked by E = −KE. The same reasoning explains why a satellite in a lower orbit always orbits faster and with a shorter period than one higher up.

ℹ️ Geostationary orbit, and weightlessness in a spacecraft

A geostationary satellite must have a period of exactly 24 hours, must orbit in the equatorial plane, and must go west to east with the Earth's rotation — all three conditions, not just the period. Putting T = 24 h into T² = 4π²r³/GM gives r ≈ 42,200 km from the centre, about 36,000 km above the surface. A polar satellite, by contrast, sits in a low orbit crossing the poles, so it sweeps a different strip on each pass and is used for imaging and weather rather than communication. Finally, astronauts float not because gravity is absent — at 400 km, g is still about 8.7 m/s² — but because they and the station are in free fall together, so the normal force between them is zero. Weightlessness is the absence of contact force, not of gravity.

Key takeaways

F = Gm₁m₂/r² and g = GM/R²; G is universal, g is a property of the planet and its position.
Going up, g ∝ 1/r²; going down, g = g(1 − d/R), falling LINEARLY to zero at the centre.
A shell exerts no net gravitational force inside itself — that is why g falls linearly below the surface.
V = −GM/r and U = −GMm/r, with the zero at infinity; negative total energy means a bound orbit.
For a satellite: KE = GMm/2r, U = −2KE, and total E = −KE = −GMm/2r.
v_orbit = √(GM/r) and v_escape = √(2gR) ≈ 11.2 km/s — a factor of exactly √2, and mass-independent.
A satellite losing energy to drag spirals inward and thereby speeds UP, since v = √(GM/r).
Geostationary needs all three: 24-hour period, equatorial plane, west-to-east, at ~36,000 km altitude.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The value of g at the centre of the Earth (assumed uniform) is:
1. Maximum
2. Zero
3. Half its surface value
4. Infinite
Show answer
Answer: B — Zero
Below the surface g = g(1 − d/R), and at the centre d = R so g = 0. Physically, a shell of matter exerts no net force inside it, so at the centre every direction is equally surrounded by mass and the pulls cancel exactly. g is maximum at the surface, not the centre.
Escape velocity from the Earth's surface is 11.2 km/s. The orbital speed of a satellite very close to the surface is about:
1. 5.6 km/s
2. 7.9 km/s
3. 11.2 km/s
4. 15.8 km/s
Show answer
Answer: B — 7.9 km/s
v_escape = √2 × v_orbit at the same radius, so v_orbit = 11.2/√2 ≈ 7.9 km/s. The tempting 5.6 comes from halving instead of dividing by √2, and 15.8 comes from multiplying by √2 in the wrong direction — escape must be the larger of the two.
The total energy of a satellite in a stable circular orbit is:
1. Positive, equal to its kinetic energy
2. Negative, equal in magnitude to its kinetic energy
3. Zero
4. Negative, equal in magnitude to its potential energy
Show answer
Answer: B — Negative, equal in magnitude to its kinetic energy
KE = GMm/2r and U = −GMm/r, so E = KE + U = −GMm/2r = −KE. The negative sign is what "bound" means: an object with E ≥ 0 escapes. Option 4 confuses the magnitudes — |U| is twice |E|, since U = −2KE.
A satellite in a low orbit gradually loses energy to atmospheric drag. Its orbital speed:
1. Decreases, since it is losing energy
2. Increases, because the orbital radius shrinks
3. Stays constant
4. Increases then decreases
Show answer
Answer: B — Increases, because the orbital radius shrinks
Losing energy makes E = −GMm/2r more negative, which requires r to decrease; and since v = √(GM/r), a smaller r gives a larger v. Potential energy falls by more than kinetic energy rises, so the total still decreases — kinetic energy up while total energy down is the relation to hold, via E = −KE.
Which is NOT a requirement for a geostationary satellite?
1. A period of 24 hours
2. An orbit in the equatorial plane
3. Motion from west to east
4. A mass equal to that of a standard communication satellite
Show answer
Answer: D — A mass equal to that of a standard communication satellite
The orbital radius depends only on the period through T² = 4π²r³/GM, and the satellite's own mass cancels out entirely — a heavy and a light satellite share the same geostationary orbit at about 36,000 km altitude. The other three conditions are all genuinely required, and a satellite meeting the period but not the equatorial plane traces a figure-of-eight rather than staying fixed.
For small h, the reduction in g at height h compared with the reduction at depth h is:
1. The same
2. Twice as large
3. Half as large
4. Four times as large
Show answer
Answer: B — Twice as large
At height h, g ≈ g(1 − 2h/R); at depth h, g = g(1 − h/R). So rising reduces g twice as fast as descending. The difference arises because above the surface the whole mass stays below you and the inverse square operates, while below it a shell contributes nothing and only the inner sphere pulls.
Astronauts float inside the International Space Station because:
1. There is no gravity at that altitude
2. They and the station are in free fall together, so the contact force is zero
3. The station spins to cancel gravity
4. Their mass becomes zero in orbit
Show answer
Answer: B — They and the station are in free fall together, so the contact force is zero
At about 400 km, g is still roughly 8.7 m/s² — gravity is very much present, and is exactly what keeps the station in orbit. Weightlessness is the absence of any normal force, because the astronaut and the floor accelerate identically. Mass is an invariant property and never becomes zero.
Two satellites orbit a planet at radii r and 4r. The ratio of their time periods T₁ : T₂ is:
1. 1 : 4
2. 1 : 8
3. 1 : 2
4. 1 : 16
Show answer
Answer: B — 1 : 8
By Kepler's third law T² ∝ r³, so T ∝ r^3/2 and the ratio is 1 : 4^3/2 = 1 : 8. The tempting 1 : 4 comes from taking T ∝ r, and 1 : 16 from squaring instead of taking the three-halves power — checking the exponent is the whole question.