Gravitational potential, denoted by Φ, is the potential energy per unit mass at a point in a gravitational field. It tells us how much energy an object would have if it were brought from infinity to that point.
Mathematically: \$\Phi = -\frac{GM}{r}\$ where G is the gravitational constant, M the mass creating the field, and r the distance from that mass.
Think of it like a hill: the higher you are, the more “potential” you have to fall down.
When an object of mass m is at a potential Φ, its gravitational potential energy is:
\$U = m\Phi = -\frac{G M m}{r}\$
Example: A 10 kg ball 10 m above Earth’s surface (Φ ≈ –9.8 m²/s²) has:
\$U = 10\,\text{kg} \times (-9.8\,\text{m}^2/\text{s}^2) = -98\,\text{J}\$
The negative sign means the energy is lower than at infinity.
When an object falls, its GPE is converted into kinetic energy. If it hits a surface, friction turns that kinetic energy into heat, raising the temperature of the surface.
Analogy: A roller‑coaster starts at the top of a hill (high potential). As it goes down, the potential energy becomes speed. When it brakes, that speed is turned into heat in the brakes.
In astrophysics, the deep gravitational potential wells of stars compress gas, heating it up to millions of kelvins – the reason stars shine!
Exam Tip: When solving for temperature change due to gravitational energy, remember the energy balance:
\$\Delta U = m c \Delta T\$
where c is the specific heat. Always check units and sign conventions.
| Parameter | Symbol | Typical Value |
|---|---|---|
| Gravitational Constant | G | 6.674×10⁻¹¹ m³ kg⁻¹ s⁻² |
| Earth’s Mass | Mₑ | 5.97×10²⁴ kg |
| Surface Potential | Φₑ | –6.26×10⁷ m² s⁻² |
Answer: Use ΔU = mgh = 2 kg × 9.8 m/s² × 50 m = 980 J. Then ΔT = ΔU/(m c) = 980 J / (1 kg × 2090 J/kg/K) ≈ 0.47 K.
Remember: Always state assumptions (e.g., no energy loss to air). Check if the problem asks for potential energy or temperature change. Use the correct sign for potential energy (negative for bound systems). Good luck! 🚀