Gravitational Potential and Orbits

SPEC 3.7.2.3, 3.7.2.4 • FIELDS

Potential energy, orbital mechanics, and escape velocity

Gravitational Potential

Definition

Gravitational potential (V) at a point is the work done per unit mass in bringing a test mass from infinity to that point.

V = - \frac{GM}{r}

Key Concept

V is ALWAYS negative

V = 0 at infinity

V more negative closer to M

Units: J kg⁻¹

Important

Don't confuse potential (V) with potential energy (Eₚ):

E_{p} = mV = - \frac{GM m}{r}

Why is Potential Negative?

The negative sign indicates that a mass is bound to the gravitational field.

At Infinity

V = 0 (zero potential)

The reference point where the mass is completely free from gravitational influence.

Near a Mass

V < 0 (negative potential)

Energy must be supplied to move the mass to infinity - it's in a "potential well".

Physical Meaning

As an object moves from infinity toward a mass, gravitational potential energy is released. Since V = 0 at infinity and energy is released, V must become negative.

Worked Example01

The gravitational potential at Earth's surface is −62.6 MJ kg⁻¹ and at the Moon's surface is −2.8 MJ kg⁻¹. Explain which body has the deeper "potential well" and what this means physically.

1

Compare magnitudes: |−62.6| > |−2.8|, so Earth's potential is more negative

2

Deeper well: Earth has the deeper potential well (more negative = deeper)

3

Physical meaning: More energy per kg is needed to escape from Earth's surface

4

To escape Earth: need to supply 62.6 MJ for each kg of mass

5

To escape Moon: need only 2.8 MJ for each kg of mass

6

Conclusion: Objects are more strongly bound to Earth - this is why rockets need more fuel to leave Earth than the Moon

Potential vs Distance

Key Feature

The gradient of the V-r graph at any point equals −g (negative of field strength).

Potential follows

V \propto \frac{1}{r}

(inverse relationship, NOT inverse square like g)

Linking Field Strength and Potential

Key Relationship

g = - \frac{d V}{d r}

Field strength equals the negative gradient of the V-r graph

g-r graph

Area Under g-r Graph

The area under the g-r graph between two points gives the change in gravitational potential (ΔV).

Equipotential Surfaces

Definition

Equipotential surfaces are surfaces where the gravitational potential has the same value at all points.

No Work Done

No work is done moving along an equipotential (ΔV = 0).

Perpendicular

Field lines are always perpendicular to equipotential surfaces.

Worked Example02

Explain why no work is done when a satellite moves along an equipotential surface, but work must be done to move it to a higher orbit.

1

Along an equipotential: All points have the same potential V, so ΔV = 0

2

Work done W = mΔV = m × 0 = zero

3

Movement is perpendicular to the gravitational field, so no work against gravity

4

To a higher orbit: Must move to a region of higher (less negative) potential

5

ΔV > 0 (potential increases), so W = mΔV > 0 → work must be done

6

This work increases the satellite's gravitational potential energy

03Worked Example

Calculate the gravitational potential at the surface of Mars and at an altitude of 500 km.

Mass of Mars = 6.42 × 10²³ kg, Radius = 3.40 × 10⁶ m

1

At surface:

V = - \frac{GM}{r} = - \frac{6.67 \times 1 0 ^{- 11} \times 6.42 \times 1 0 ^{23}}{3.40 \times 1 0 ^{6}}

2

V_{s u r f a ce} = - 1.26 \times 1 0^{7} J kg^{- 1}

3

At 500 km: r = 3.40 × 10⁶ + 0.50 × 10⁶ = 3.90 × 10⁶ m

4

V_{500 km} = - \frac{4.28 \times 1 0 ^{13}}{3.90 \times 1 0 ^{6}} = - 1.10 \times 1 0^{7} J kg^{- 1}

Note: Potential is less negative (higher) at greater altitude.

Worked Example04

The gravitational potential at the surface of a planet of radius R is V₀. At what height above the surface is the potential equal to V₀/3?

1

At surface (r = R):

V_{0} = - \frac{GM}{R}

2

At height h (r = R + h):

V = - \frac{GM}{R + h} = \frac{V _{0}}{3}

3

Using

V \propto \frac{1}{r}

: ratio

\frac{V _{0}}{V} = \frac{R + h}{R}

4

Substituting:

\frac{V _{0}}{V _{0} /3} = \frac{R + h}{R}

→

3 = \frac{R + h}{R}

5

Solving: 3R = R + h →

h = 2R

Key insight: To reduce |V| by factor of 3, distance from centre must increase by factor of 3.

Work Done and Potential Difference

Key Formula

The work done moving mass m through potential difference ΔV:

Δ W = m Δ V

Δ E_{p} = m (V_{2} - V_{1})

Δ E_{p} = GM m (\frac{1}{r _{1}} - \frac{1}{r _{2}})

Sign Convention

ΔW > 0: Work done ON the mass (moving away from M)

ΔW < 0: Work done BY the field (mass falling toward M)

05Worked Example

A 500 kg satellite is moved from altitude 400 km to 800 km above Earth. Calculate the work done.

M_Earth = 5.97 × 10²⁴ kg, R_Earth = 6.37 × 10⁶ m

1

r₁ = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m

2

r₂ = 6.37 × 10⁶ + 0.80 × 10⁶ = 7.17 × 10⁶ m

3

V_{1} = - \frac{3.98 \times 1 0 ^{14}}{6.77 \times 1 0 ^{6}} = - 5.88 \times 1 0^{7} J kg^{- 1}

4

V_{2} = - \frac{3.98 \times 1 0 ^{14}}{7.17 \times 1 0 ^{6}} = - 5.55 \times 1 0^{7} J kg^{- 1}

5

ΔW = 500 × [(-5.55 × 10⁷) - (-5.88 × 10⁷)] = 1.65 × 10⁹ J

Orbital Speed

For a circular orbit, gravitational force provides the centripetal force:

Step 1: Equate Forces

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

Gravitational force = Centripetal force

Step 2: Cancel m and one r

\frac{GM}{r} = v^{2}

Step 3: Solve for v

v = \frac{GM}{r}

F (gravity) provides centripetal force

Key Points

Orbital speed decreases with increasing r • Speed is independent of satellite mass m • Also:

v = g r

Orbital Period - Kepler's Third Law

Derivation

T = \frac{2 π r}{v} = 2 π \frac{r ^{3}}{GM}

Kepler's Third Law

Squaring gives:

T^{2} = \frac{4 π ^{2}}{GM} r^{3}

This shows T² ∝ r³ - period squared is proportional to radius cubed.

Comparing Orbits

For satellites orbiting the same body:

\frac{T _{1}^{2}}{T _{2}^{2}} = \frac{r _{1}^{3}}{r _{2}^{3}}

Worked Example06

Explain why satellites in higher orbits take longer to complete one orbit. Your answer should address both the distance travelled and the orbital speed.

1

Distance factor: Circumference = 2πr, so larger r means greater distance to travel

2

Speed factor: v = √(GM/r), so larger r means lower orbital speed

3

Combined effect: T = distance/speed = 2πr/v

4

As r increases: numerator (2πr) increases AND denominator (v) decreases

5

Double effect: Both factors work together to increase T

6

This explains why T² ∝ r³ (not just T ∝ r): the period increases faster than the radius because both distance AND reduced speed contribute

07Worked Example

The ISS orbits at altitude 400 km. Calculate its orbital speed and period.

M_Earth = 5.97 × 10²⁴ kg, R_Earth = 6.37 × 10⁶ m

1

r = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m

2

v = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24}}{6.77 \times 1 0 ^{6}} = 5.88 \times 1 0^{7}

3

v = 7670 m s^{- 1}

(7.67 km/s)

4

T = \frac{2 π \times 6.77 \times 1 0 ^{6}}{7670} = 5540 s = 92.4 minutes

Energy of a Satellite

Kinetic Energy

K E = \frac{1}{2} m v^{2} = \frac{GM m}{2 r}

Gravitational Potential Energy

GPE = - \frac{GM m}{r}

Total Energy

E_{t o t a l} = K E + GPE = - \frac{GM m}{2 r}

Key Relationships

Total energy is negative → satellite is bound to orbit

|GPE| = 2 × KE → magnitude of GPE is twice the KE

Total = −KE = ½GPE

Worked Example08

When a satellite loses energy due to atmospheric drag, it drops to a lower orbit and speeds up. This seems to contradict conservation of energy. Explain why this happens.

1

Energy lost to drag: Friction removes energy from the satellite

2

Satellite drops: With less total energy, it falls to a lower orbit (smaller r)

3

At lower orbit: v = √(GM/r) — smaller r means larger v!

4

Energy accounting: KE = GMm/2r increases (r smaller), but GPE = −GMm/r becomes more negative (larger magnitude)

5

Net effect: ΔGPE = −2 × ΔKE, so when GPE drops by 100 J, KE rises by 50 J

6

Conclusion: Total energy decreases (50 J lost to drag), but the satellite speeds up because KE increases at the expense of GPE

Energy in Different Orbits

Circular Orbits

Satellite stays on same equipotential

GPE constant, KE constant

Speed constant throughout

Total energy constant

Elliptical Orbits

Moves between equipotentials

GPE varies, KE varies

Faster at perigee, slower at apogee

Total energy still constant

Exam Tip

In elliptical orbits, KE and GPE interchange but total mechanical energy is conserved (no air resistance in space).

Worked Example09

A satellite in an elliptical orbit moves faster at perigee (closest point) than at apogee (furthest point). Explain why, in terms of energy.

1

Total energy is constant: E = KE + GPE throughout the orbit

2

At perigee (r small): GPE = −GMm/r is more negative (lower)

3

Since E is constant, if GPE decreases, KE must increase

4

Higher KE means higher speed → satellite moves fastest at perigee

5

At apogee (r large): GPE is less negative (higher), so KE is lower

6

Lower KE means lower speed → satellite moves slowest at apogee

Key principle: Conservation of energy causes the speed variation in elliptical orbits.

10

Worked Example

A 200 kg satellite orbits Earth at radius 8.00 × 10⁶ m. Calculate KE, GPE, and total energy.

1

K E = \frac{GM m}{2 r} = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24} \times 200}{2 \times 8.00 \times 1 0 ^{6}}

2

KE = 4.98 × 10⁹ J

3

GPE = - \frac{GM m}{r} = - 9.96 \times 1 0^{9} J

4

Total = 4.98 × 10⁹ − 9.96 × 10⁹ = −4.98 × 10⁹ J

Check: |KE| = ½|GPE| ✓ and Total = −KE ✓

Escape Velocity

Definition

Escape velocity is the minimum speed needed to escape a gravitational field completely (reach infinity with zero velocity).

Condition: Total energy = 0 (just enough to reach infinity)

At escape: KE + GPE = 0 → KE = |GPE|

Derivation

Setting KE equal to |GPE|:

\frac{1}{2} m v_{esc}^{2} = \frac{GM m}{r}

Solving for v_esc:

v_{esc} = \frac{2 GM}{r}

v_{esc} = 2 g R

v_{esc} = 2 \times v_{or bi t a l}

Key Point

Escape velocity is independent of the mass of the escaping object - it only depends on M and r.

11

Worked Example

A spacecraft (mass 10,000 kg) and a small rock (mass 1 kg) are both launched from Earth's surface. Explain why they both need the same speed to escape Earth's gravity.

1

For the spacecraft: KE = ½(10000)v² and GPE = −GM(10000)/R

2

For the rock: KE = ½(1)v² and GPE = −GM(1)/R

3

Key insight: Both KE and GPE are proportional to the object's mass m

4

Setting KE = |GPE|: ½mv² = GMm/R

5

The mass m cancels: ½v² = GM/R → v = √(2GM/R)

6

Conclusion: Escape velocity depends only on the planet (M and R), not on the escaping object's mass. A feather and a rocket need the same speed!

12

Worked Example

(a) Calculate Earth's escape velocity from the surface.

(b) Show this equals √2 × orbital velocity.

g₀ = 9.81 m s⁻², R = 6.37 × 10⁶ m

1

(a)

v_{esc} = 2 g R = 2 \times 9.81 \times 6.37 \times 1 0^{6}

2

v_{esc} = 1.25 \times 1 0^{8} = 11, 200 m s^{- 1}

= 11.2 km/s

3

(b)

v_{or b} = g R = 9.81 \times 6.37 \times 1 0^{6} = 7910 m s^{- 1}

4

\frac{v _{esc}}{v _{or b}} = \frac{11200}{7910} = 1.41 = 2

✓

Synchronous vs Geostationary

Geosynchronous

Period T = 24 hours

Can be in any orbital plane

Passes same point at same time daily

Does NOT stay fixed above one point

Geostationary

Period T = 24 hours

Must be above equator

Orbits west to east

Remains fixed above one point

Geostationary Requirements

T = 24 hours exactly

Equatorial orbit (above equator)

Circular orbit

Same direction as Earth's rotation (west to east)

Low-Orbit vs Geostationary

Low Earth Orbit (LEO)

Altitude: 200-2000 km

Period: ~90 minutes

Speed: ~7.5 km/s (faster)

Better imaging (closer)

Multiple satellites needed

Uses: ISS, weather, imaging, Starlink

Geostationary (GEO)

Altitude: ~35,800 km

Period: 24 hours

Speed: ~3 km/s (slower)

Fixed position coverage

Single satellite covers large area

Uses: TV, communications, weather

Common Error

Using altitude instead of radius: r = R + h (always add Earth's radius!)

13

Worked Example

Calculate the radius and altitude of a geostationary orbit.

M_Earth = 5.97 × 10²⁴ kg, R_Earth = 6.37 × 10⁶ m

1

T = 24 × 60 × 60 = 86,400 s

2

From T² = 4π²r³/GM:

r^{3} = \frac{GM T ^{2}}{4 π ^{2}}

3

r^{3} = \frac{6.67 \times 1 0 ^{- 11} \times 5.97 \times 1 0 ^{24} \times ( 86400 ) ^{2}}{4 π ^{2}} = 7.53 \times 1 0^{22}

4

r = ∛(7.53 × 10²²) = 4.22 × 10⁷ m = 42,200 km

5

Altitude h = 42,200 − 6,370 = 35,800 km

Practice

Question 1: Calculate the gravitational potential at Earth's surface and at altitude 2R. Find the work done per kg to move between these positions.

At surface (r = R): V₁ = −GM/R = −6.26 × 10⁷ J kg⁻¹

At altitude 2R (r = 3R): V₂ = −GM/3R = −2.09 × 10⁷ J kg⁻¹

Work per kg: ΔV = (−2.09 × 10⁷) − (−6.26 × 10⁷) = 4.17 × 10⁷ J kg⁻¹

Practice

Question 2: A satellite has period T at radius r. If moved to radius 4r, what is its new period?

Kepler's Third Law: T² ∝ r³

\frac{T _{2}^{2}}{T _{1}^{2}} = \frac{( 4 r ) ^{3}}{r ^{3}} = 64

T_{2} = 64 \times T = 8T

Practice

Question 3: A satellite orbits with total energy E (negative). How much additional energy is needed to escape?

Total energy: E = −GMm/2r (negative, bound)

Kinetic energy: KE = GMm/2r = |E|

To escape: Need total energy = 0

Energy required = 0 − E = −E = |E| = KE of the satellite

Summary

Key Formulas

QuantityFormula

Gravitational potential

V = - \frac{GM}{r}

Field from potential

g = - \frac{d V}{d r}

Work done

Δ W = m Δ V

Orbital speed

v = \frac{GM}{r}

Kepler's Third Law

T^{2} = \frac{4 π ^{2}}{GM} r^{3}

Total orbital energy

E = - \frac{GM m}{2 r}

Escape velocity

v_{esc} = \frac{2 GM}{r} = 2 \times v_{or b}

Key Exam Points

V is always negative; g = −dV/dr; area under g-r graph = ΔV

For orbits: |KE| = ½|GPE|, Total E = −KE

Geostationary: T = 24h, equatorial, r ≈ 42,200 km