Interactive physics simulator
Gravitational Potential (V = -GM/r)
Investigate gravitational potential fields, potential energy wells, and escape velocities. Launch satellites to test stable circular orbits, study the combined potential barrier of binary planetary co-orbitals, and track kinetic/potential energy transformations during vertical rocket launches.
Gravitational Potential Lab
Configure physics parameters, select celestial bodies, and drag bobs/probes to observe gravitational potential values.
Live Telemetry
- Planet Mass (M)
- 1.00 M⊕
- Planet Radius (R)
- 6,371 km
- Probe Distance (r)
- 12,500 km
- Potential (V)
- -31.9 MJ/kg
- Escape Velocity (v_esc)
- 7.99 km/s
- Orbital Speed (v_orb)
- 5.65 km/s
- Current Speed (v)
- 0.00 km/s
Understanding Gravitational Potential
Gravitational Potential (represented by V) at a point in space is a scalar quantity that describes the potential energy stored per unit mass at that position within a gravitational field. It represents the work done against gravity to move a test mass from an infinite distance (where gravity is zero) to that specific coordinate.
Because gravity is always an attractive force, bringing a mass closer to another mass releases energy. Defining potential as zero at infinity (reference reference V = 0 at r = ∞) means potential values are negative everywhere else. For a spherical planetary body of mass M, the potential at a distance r from its center is:
Where:
• G is the universal gravitational constant (6.674 × 10⁻¹¹ N·m²/kg²)
• M is the mass of the planetary body in kilograms (kg)
• r is the distance from the center of mass in meters (m)
Unlike gravitational field strength ($g$, a vector), potential is a scalar, meaning fields from multiple objects simply add up directly: Vnet = V₁ + V₂ + V₃...
Interactive Potential Calculator
Adjust parameters to calculate local potential and escape velocity:
Planetary Potential & Escape Values
Compare surface properties across celestial bodies:
| Celestial Body | Mass (M) | Radius (R) | Surface Potential (V₀) | Escape speed (v_esc) |
|---|---|---|---|---|
| Moon | 0.012 M⊕ | 1,737 km | -2.8 MJ/kg | 2.38 km/s |
| Mars | 0.107 M⊕ | 3,390 km | -12.6 MJ/kg | 5.02 km/s |
| Venus | 0.815 M⊕ | 6,052 km | -53.7 MJ/kg | 10.36 km/s |
| Earth | 1.000 M⊕ | 6,371 km | -62.6 MJ/kg | 11.19 km/s |
| Jupiter | 317.8 M⊕ | 69,911 km | -1,812.2 MJ/kg | 59.54 km/s |
Math & Energy Profiles
How gravitational potential drives orbital energy:
- Potential Well: As a probe falls closer (r decreases), the potential well deepens. Escape speed scales as: vesc = √(−2V) = √(2GM/r).
- Total Energy:
• Bound Orbit: E = KE + PE < 0. The projectile has insufficient speed to escape the well.
• Escaping Orbit: E = KE + PE ≥ 0. The projectile breaks free. - Binary Saddle Peak: In Earth-Moon systems, the potential reaches a local maximum barrier at L1. Probes require launch velocities surmounting this barrier: vtransit ≥ √(2|Vlaunch - VL1|).
Single Potential Well Curve
Visual representation of the negative potential well V = -GM/r centered around a planet:
Solved Examples
Calculate the gravitational potential at the surface of Venus, given that Venus has a mass of 4.87 × 10²⁴ kg and a radius of 6,052 km. Use G = 6.674 × 10⁻¹¹ N·m²/kg².
- Identify the given values: mass (M) = 4.87 × 10²⁴ kg, radius (R) = 6,052 km = 6.052 × 10⁶ m, G = 6.674 × 10⁻¹¹ N·m²/kg².
- Recall the formula for gravitational potential of a sphere: V = -G·M / R.
- Substitute the values: V = -(6.674 × 10⁻¹¹) * (4.87 × 10²⁴) / (6.052 × 10⁶).
- Calculate the numerator: G * M = 3.2502 * 10¹⁴ m³/s².
- Divide by the radius: V = -3.2502 * 10¹⁴ / (6.052 × 10⁶) = -53,705,188 J/kg.
- Convert to megajoules: V = -53.71 MJ/kg.
- Verify: The potential is negative, meaning work is released when a unit mass falls from infinity to Venus's surface.
Answer: V = -53.7 MJ/kg
Determine the escape velocity of a satellite orbiting Earth at an altitude of 3,629 km. Earth's mass is 5.97 × 10²⁴ kg and its radius is 6,371 km. Use G = 6.674 × 10⁻¹¹ N·m²/kg².
- Identify the total distance (r) from Earth's center: r = Radius + Altitude = 6,371 km + 3,629 km = 10,000 km = 1.0 × 10⁷ m.
- Identify the given values: M = 5.972 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N·m²/kg².
- Recall the formula for escape velocity relative to potential: v_esc = sqrt(-2·V) = sqrt(2·G·M / r).
- Substitute the values: v_esc = sqrt(2 * 6.674 × 10⁻¹¹ * 5.972 × 10²⁴ / 1.0 × 10⁷).
- Calculate the value inside the square root: 2 * 3.9857 × 10¹⁴ / 1.0 × 10⁷ = 7.9714 × 10⁷ m²/s².
- Take the square root: v_esc = sqrt(7.9714 × 10⁷) ≈ 8,928 m/s = 8.93 km/s.
- Verify: This escape velocity is lower than Earth's surface escape velocity (11.2 km/s), which is correct because potential weakens with altitude.
Answer: v_esc = 8.93 km/s
A rocket launches vertically from Mars (mass = 6.39 × 10²³ kg, radius = 3,390 km) with a launch speed equal to 120% of its surface escape speed. Calculate its residual speed when it escapes to a very large distance (infinity) from Mars.
- Identify the given values: Mars mass (M) = 6.39 × 10²³ kg, radius (R) = 3.39 × 10⁶ m.
- Calculate surface escape velocity: v_esc = sqrt(2·G·M / R) = sqrt(2 * 6.674 × 10⁻¹¹ * 6.39 × 10²³ / 3.39 × 10⁶) ≈ 5,017 m/s = 5.02 km/s.
- Identify launch speed: v_init = 1.20 * v_esc ≈ 6,020 m/s.
- Use conservation of energy between launch and infinity: E_total = KE_init + PE_init = KE_inf + PE_inf.
- Recall that potential at infinity is zero (PE_inf = 0), so: 1/2 * m * v_init² - G·M·m/R = 1/2 * m * v_inf².
- Notice that -G·M/R is equal to -1/2 * v_esc², so: 1/2 * m * v_init² - 1/2 * m * v_esc² = 1/2 * m * v_inf².
- Cancel out 1/2 * m: v_inf² = v_init² - v_esc².
- Substitute the launch speed expression (v_init = 1.2 * v_esc): v_inf² = (1.2 * v_esc)² - v_esc² = (1.44 - 1.0) * v_esc² = 0.44 * v_esc².
- Calculate final speed: v_inf = sqrt(0.44) * v_esc = 0.663 * 5,017 m/s ≈ 3,328 m/s = 3.33 km/s.
- Verify: When launching faster than escape velocity, the rocket escapes gravity with positive kinetic energy, leaving it with excess speed.
Answer: v_inf = 3.33 km/s
Common Mistakes
- "Potential and potential energy are the same": False. Gravitational potential (V) is a property of the field itself (measured in J/kg). Gravitational potential energy (U) depends on the mass of the probe object (measured in Joules), where U = mV.
- "Negative sign means negative values are physically meaningless": False. The negative sign is a mathematical representation of attraction. Because zero potential is chosen at infinity, potential is negative everywhere else, indicating a bound energy state.
- "Escape velocity requires the rocket to travel at that speed the entire trip": False. Escape velocity is the initial launch velocity needed to coast to infinity without any further fuel. A powered rocket can climb at any speed as long as it has fuel.
- "Gravity is zero at Lagrange points, so potential is zero": False. At the L1 saddle point, the net gravitational force is zero (g = 0), but the combined potential (V) is at its most positive relative local peak while still being deeply negative in absolute terms.
Practice Questions
1. Explain the physical significance of the negative sign in the gravitational potential equation V = -GM/r. What would a positive potential represent?
The negative sign represents that the gravitational force is attractive. By defining the zero potential reference point at infinity (r = infinity), any position closer than infinity requires work to escape, meaning potential energy must be added to reach zero. Thus, the potential is negative everywhere. A positive potential would imply a repulsive force, where energy is released as objects move apart.
2. If gravitational potential is a scalar quantity, why do we use vector arrows in gravitational fields? Compare how potential and field strength are related.
Gravitational field strength (g) is a vector that points in the direction of the gravitational force. Gravitational potential (V) is a scalar indicating the potential energy per unit mass. They are related mathematically: the field strength is the negative gradient of the potential (g = -dV/dr). Visually, field vector arrows point "downhill" perpendicular to equipotential contour rings, from high potential to low potential.
3. A satellite in a circular orbit of radius r has a speed of v_orb. If the satellite needs to escape Earth entirely, by what percentage must its speed be increased? Show the relation between orbital and escape velocity.
For a stable circular orbit, orbital speed is v_orb = sqrt(GM/r). The escape speed at that distance is v_esc = sqrt(2GM/r) = sqrt(2) * v_orb ≈ 1.414 * v_orb. To escape, the speed must increase by a factor of sqrt(2) - 1 ≈ 0.414, or exactly 41.4%.
4. In a binary planetary system (e.g. Earth and Moon), explain why the L1 Lagrange point is considered a potential saddle point. Is a satellite at L1 stable or unstable?
Along the line connecting the Earth and the Moon, the potential rises to a local peak at L1 because the negative potential wells pull in opposite directions. However, perpendicular to this line, L1 is a local potential valley. This makes L1 a saddle point. A satellite at L1 is unstable: if perturbed slightly toward either the Earth or the Moon, it will fall down the potential well, escaping L1.
FAQ
Frequently Asked Questions
What is gravitational potential?
Gravitational potential (represented by V) at a point in a gravitational field is the work done per unit mass in bringing a small test mass from infinity to that point. It is calculated as V = -G*M/r.
What are the SI units of gravitational potential?
The SI unit of gravitational potential is Joules per kilogram (J/kg).
Why is gravitational potential always negative?
By convention, gravitational potential is defined as zero at infinity (infinitely far from any mass). Since gravity is an attractive force, work is done by the field rather than against it as a mass moves closer, making the potential energy and potential negative everywhere closer than infinity.
Is gravitational potential a vector or a scalar?
It is a scalar quantity. Unlike the gravitational field strength (g), which is a vector pointing in the direction of the force, gravitational potential (V) has only magnitude and no direction.
How does gravitational potential relate to gravitational potential energy?
Gravitational potential (V) is the potential energy per unit mass. The gravitational potential energy (U) of a mass (m) is equal to the mass multiplied by the potential at that point: U = m*V.
What is a gravitational potential well?
A potential well is a conceptual model representing the gravitational potential field around a massive body. Plotted in 2D or 3D, it looks like a deep funnel or well centered on the mass. The closer an object gets, the deeper it falls into the well (more negative V), requiring energy to climb back out.
What is escape velocity?
Escape velocity is the minimum speed an object must possess to break free from a body's gravitational field without further propulsion. It is calculated as v_esc = sqrt(2*G*M/R) or v_esc = sqrt(-2*V).
What is orbital velocity?
Orbital velocity is the speed needed to maintain a stable circular orbit around a celestial body. For a circular orbit of radius r, it is v_orb = sqrt(G*M/r), which is exactly equal to v_esc / sqrt(2).
What is the significance of the potential barrier at L1 in a binary system?
In a binary system (like Earth-Moon), the L1 Lagrange point is located at the local maximum of the potential profile along the axis between the bodies. This point acts as an unstable saddle barrier. A spacecraft traveling from Earth to the Moon must have enough kinetic energy to clear this potential peak to fall into the Moon's gravity well.
How does gravitational potential vary inside a planet of uniform density?
Inside a uniform spherical planet, the potential does not follow the -1/r shape. Instead, it follows a parabolic shape: V(r) = -G*M*(3*R^2 - r^2) / (2*R^3). At the exact center (r = 0), the potential reaches its most negative finite value of -1.5*G*M/R.
Who first introduced the mathematical concept of potential?
The concept of potential was first introduced by the French mathematician Joseph-Louis Lagrange in 1773, and later developed extensively by Pierre-Simon Laplace and George Green (who coined the term 'potential function').