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- Introduction
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Introduction
Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. When we do work against the Earth's gravitational force to move an object away from Earth, this work is stored as potential energy in the system. The concept of potential energy helps us understand energy conservation and predict the motion of objects under gravity. Unlike other forms of energy, potential energy has a relative value, and we choose a reference point (usually at infinity) where potential energy is zero. This allows us to calculate how potential energy changes as objects move within a gravitational field.
Formula: Gravitational Potential Energy
U(r) = -\[\frac {GMm}{r}\]
Where:
- U(r) = Gravitational potential energy at distance r from Earth's center
- G = Universal gravitational constant (6.67 × 10⁻¹¹ N·m²/kg²)
- M = Mass of Earth (kg)
- m = Mass of the object (kg)
- r = Distance between the centers of mass of Earth and object (m)
- Negative sign = Shows that potential energy is negative (zero at infinity)
Derivation
1. Work Done Against Gravitational Force
- When we move an object against Earth's gravitational force, we do work on the system
- This work appears as an increase in potential energy
- The relationship is: dU = -\[\vec F_g\] · d\[\vec r\]
- The negative sign indicates that work done by us (external agent) is opposite to the gravitational force
2. Change in Potential Energy (Integration)
-
To find total change in potential energy when moving from position riri to rf:
ΔU = \[\int_{r_i}^{r_f}dU=\int_{r_i}^{r_f}(-\vec{F}_g\cdot d\vec{r})\]
3. Gravitational Force Expression
-
Earth's gravitational force on mass m:
\[\vec{F}_g=-\frac{GMm}{r^2}\hat{r}\]
-
The negative sign appears because:
-
\[\hat r\] is the unit vector pointing away from Earth's center (outward)
-
\[\vec F_g\] is directed toward Earth's center (inward)
4. Substituting and Solving the Integral
ΔU = \[\int_{r_i}^{r_f}\left(-\left(-\frac{GMm}{r^2}\hat{r}\right)\right)\cdot d\vec{r}\]
= GMm\[\int_{r_i}^{r_f}\frac{dr}{r^2}\]
(Note: d\[\vec r\] is along \[\hat r\], so \[\hat r\] · d\[\vec r\] )
= GMm\[\left[-\frac{1}{r}\right]_{r_i}^{r_f}=GMm\left(\frac{1}{r_i}-\frac{1}{r_f}\right)\]
5. Choosing the Reference Point
- For potential energy, the absolute value is not defined—only changes matter
- We conventionally choose zero potential energy at infinity (r = ∞)
- This makes calculations logical and convenient
6. Final Formula Derivation
- Setting initial point: ri = ∞ where U(∞) = 0
- Final point: rf = r (distance where we calculate potential energy)
U(r) = \[GMm\left(\frac{1}{\infty}-\frac{1}{r}\right)=-\frac{GMm}{r}\]
Example
Problem: What will be the change in potential energy of a body of mass m when it is raised from height RE above Earth's surface to \[\frac {5}{2}\]RE above Earth's surface? RE and ME are the radius and mass of Earth respectively.
Solution:
1. Identify distances from Earth's center
- Initial position: height = RE above surface → distance from center = ri =RE + RE =
- Final position: height = \[\frac {4}{2}\]RE above surface → distance from center = rf = RE + \[\frac {5}{2}\]RE = \[\frac {7}{2}\]RE = 3.5RE
Apply the change in potential energy formula
ΔU = GMm\[\left(\frac{1}{r_i}-\frac{1}{r_f}\right)\]
Substitute the values
ΔU = GmME\[\left(\frac{1}{2R_E}-\frac{1}{3.5R_E}\right)\]
Simplify
ΔU = \[\frac{GmM_E}{R_E}\left(\frac{1}{2}-\frac{1}{3.5}\right)\]
= \[\frac{GmM_E}{R_E}\left(\frac{1}{2}-\frac{2}{7}\right)\]
= \[\frac{GmM_{E}}{R_{E}}\left(\frac{7-4}{14}\right)=\frac{GmM_{E}}{R_{E}}\times\frac{3}{14}\]
Final answer
\[\Delta U\approx0.214\frac{GmM_E}{R_E}\mathrm{~or~}2.14\frac{GmM_E}{R_E}\] (in simplified fractional form)
Interpretation: The positive change indicates that potential energy increases when the object is raised away from Earth's surface, requiring external work to be done against the gravitational force.
