# Gravitational Potential Energy ## What is Gravitational Potential Energy? - Gravitational potential energy ($E_p$) is the energy stored in a mass due to its **position** in a gravitational field - Objects gain $E_p$ when they are raised to a higher position against gravity - Objects lose $E_p$ when they fall to a lower position

Gravitational potential energy is the energy stored in an object due to its height above a reference point in a gravitational field

- When you carry a bag of shopping up the stairs, you transfer energy to the bag, increasing its gravitational potential energy - When a book falls off a shelf, its gravitational potential energy decreases as it drops ## The GPE Equation - For objects near Earth's surface, where the gravitational field is **uniform**, gravitational potential energy is calculated using: $$\Delta E_p = mg\Delta h$$ - Where: - $\Delta E_p$ = change in gravitational potential energy (J) - $m$ = mass of the object (kg) - $g$ = gravitational field strength (9.81 N kg$^{-1}$ on Earth) - $\Delta h$ = change in vertical height (m) - The symbol $\Delta$ (delta) represents "change in" - Ground level is typically taken as the zero reference point for GPE ### Key Points About the Equation - GPE depends on three factors: **mass**, **gravitational field strength**, and **height** - Doubling the mass doubles the GPE - Doubling the height doubles the GPE - The equation only applies where the gravitational field is uniform (such as near Earth's surface) - At greater distances from Earth (such as in space), different equations are needed ### Visualizing the GPE-Height Relationship The graph below shows how gravitational potential energy increases linearly with height for an object of mass 10 kg:

- The straight-line graph confirms the **direct proportionality** between GPE and height - The gradient of the line equals $mg$ (mass × gravitational field strength) - For different masses, the line would have a different gradient but remain straight ## Deriving the GPE Equation - The GPE equation can be derived from the concept of work done against gravity - When an object is lifted vertically, work must be done against the gravitational force ### Step-by-Step Derivation 1. **Consider a mass $m$ being lifted through a vertical height $\Delta h$** - The weight of the object is $W = mg$ (where $g$ is gravitational field strength) - This weight acts downwards throughout the lift 2. **Calculate the work done lifting the mass** - Work done = force × distance moved in direction of force - The upward force needed equals the weight: $F = mg$ - Distance moved vertically = $\Delta h$ - Therefore: $W = F \times \Delta h = mg \times \Delta h$ 3. **Apply the energy transfer principle** - Work done on the object equals the energy transferred to it - This energy is stored as gravitational potential energy - Therefore: $\Delta E_p = mg\Delta h$ 4. **Setting the reference point** - We typically consider the mass to have zero potential energy at ground level - This gives us the final equation: $\Delta E_p = mg\Delta h$

The derivation shows that GPE is fundamentally the work done against gravity when lifting an object

## Worked Example: Crane Lifting Concrete **Question:** A construction crane lifts a concrete slab of mass 850 kg through a vertical distance of 12 m. Calculate the gain in gravitational potential energy of the slab. **Solution:** | Quantity | Value | Unit | |----------|-------|------| | Mass, $m$ | 850 | kg | | Gravitational field strength, $g$ | 9.81 | N kg$^{-1}$ | | Height change, $\Delta h$ | 12 | m | **Step 1:** Write down the GPE equation $$\Delta E_p = mg\Delta h$$ **Step 2:** Substitute the values $$\Delta E_p = 850 \times 9.81 \times 12$$ **Step 3:** Calculate the result $$\Delta E_p = 100,062 \text{ J} = 1.0 \times 10^5 \text{ J (2 s.f.)}$$ The concrete slab gains **100 kJ** of gravitational potential energy. ## Energy Exchanges: GPE and Kinetic Energy - In many real situations, gravitational potential energy is converted to kinetic energy, or vice versa - Common examples include: - A tennis ball thrown upward (KE → GPE) - A rollercoaster descending a hill (GPE → KE) - A diver jumping off a diving board (GPE → KE) - A bungee jumper at the top of a bridge (GPE → KE → elastic PE) ### Conservation Principle - Assuming **no energy is lost** to resistive forces (like air resistance or friction): $$\text{Loss in GPE} = \text{Gain in KE}$$ Or equivalently: $$mg\Delta h = \frac{1}{2}mv^2$$ - Notice that mass $m$ appears on both sides, so it can cancel out when solving for velocity - This principle allows us to calculate final speeds, maximum heights, and other useful quantities

The equation $\Delta E_p = mg\Delta h$ only works near Earth's surface where $g$ is approximately constant

### Energy Conversion During a Fall The graph below illustrates how energy is converted from GPE to KE as an object falls from a height of 30 m:

- The GPE line decreases as the object falls (losing height) - The KE line increases as the object accelerates (gaining speed) - At any point during the fall: **Total Energy = GPE + KE = constant** (2943 J in this example) - The two lines are mirror images, demonstrating perfect energy conservation ## Worked Example: Cyclist Descending a Hill **Question:** A cyclist with a total mass (including bike) of 68 kg freewheels down a hill. She descends through a vertical height of 35 m. Assuming all GPE is converted to kinetic energy and there are no resistive forces, calculate her speed at the bottom of the hill. **Solution:** **Step 1:** Identify the energy transfer - At the top: maximum GPE, zero KE (starting from rest) - At the bottom: zero GPE (reference point), maximum KE **Step 2:** Apply conservation of energy $$\text{GPE lost} = \text{KE gained}$$ $$mg\Delta h = \frac{1}{2}mv^2$$ **Step 3:** Cancel mass and rearrange for $v$ $$g\Delta h = \frac{1}{2}v^2$$ $$v^2 = 2g\Delta h$$ $$v = \sqrt{2g\Delta h}$$ **Step 4:** Substitute values $$v = \sqrt{2 \times 9.81 \times 35}$$ $$v = \sqrt{686.7}$$ $$v = 26.2 \text{ m s}^{-1}$$ The cyclist's speed at the bottom is **26 m s$^{-1}$**. ## Energy Losses in Real Situations - In practice, not all GPE is converted to useful kinetic energy - Some energy is **dissipated** (wasted) as: - Thermal energy (heat) due to friction or air resistance - Sound energy - Deformation of materials ### Accounting for Energy Losses - When energy is lost to resistive forces: $$\text{Loss in GPE} = \text{Gain in KE} + \text{Work done against resistive forces}$$ - If a question states that a certain percentage of GPE is lost, you must account for this: $$\text{Useful KE} = \text{Total GPE} \times \left(\frac{\text{percentage retained}}{100}\right)$$ ## Worked Example: Parachutist with Air Resistance **Question:** A parachutist of mass 75 kg jumps from a hot air balloon at a height of 240 m. Due to air resistance, only 65% of her initial GPE is converted to kinetic energy when she reaches the ground. Calculate her speed on landing. **Solution:** **Step 1:** Calculate the initial GPE $$\Delta E_p = mg\Delta h = 75 \times 9.81 \times 240 = 176,580 \text{ J}$$ **Step 2:** Calculate the kinetic energy gained $$\text{KE gained} = 0.65 \times 176,580 = 114,777 \text{ J}$$ **Step 3:** Use the kinetic energy equation $$\frac{1}{2}mv^2 = 114,777$$ **Step 4:** Rearrange and solve for $v$ $$v^2 = \frac{2 \times 114,777}{75} = 3,060.72$$ $$v = \sqrt{3,060.72} = 55.3 \text{ m s}^{-1}$$ The parachutist's landing speed is **55 m s$^{-1}$**.

When calculating with percentages, convert to decimals by dividing by 100 (e.g., 65% = 0.65)

## Situations Involving Angles - When objects move on inclined surfaces, only the **vertical component** of the displacement matters for GPE - For a distance $d$ along a slope at angle $\theta$ to the horizontal: $$\Delta h = d \sin\theta$$ - This vertical height change is then used in the GPE equation ### Example: Object on a Ramp - A box slides down a 5.0 m ramp inclined at 30° to the horizontal - Vertical height change: $\Delta h = 5.0 \times \sin(30°) = 5.0 \times 0.5 = 2.5$ m - Only this 2.5 m matters for calculating the change in GPE ## Summary Table | Scenario | GPE Change | Energy Transfer | |----------|------------|-----------------| | Lifting an object | Increases | Work done → GPE stored | | Object falling freely | Decreases | GPE → KE | | Object thrown upward | Increases then decreases | KE → GPE → KE | | Object on slope | Decreases (moving down) | GPE → KE (+ work against friction) | ## Key Reminders - GPE is measured in joules (J) - The reference point for zero GPE is arbitrary but typically ground level - The $mg\Delta h$ equation assumes constant $g$ (uniform field) - Always use the **vertical** height change, not distance along a slope - In real situations, some energy is always dissipated to surroundings - Mass cancels when equating GPE and KE, simplifying calculations