Gravitational potential and energy IB DP Physics Study Notes - 2025 Syllabus
Gravitational potential and energy IB DP Physics Study Notes
Gravitational potential and energy IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
that the gravitational potential energy \(E_p\) of a system is the work done to assemble the system from infinite separation of the components of the system
the gravitational potential energy for a two-body system as given by \(E_p = -G\frac{m_1m_2}{r}\) where r is the separation between the centre of mass of the two bodies
that the gravitational potential \(V_g\) at a point is the work done per unit mass in bringing a mass from infinity to that point as given by \(V_g = -G\frac{M}{r}\)
Standard level and higher level: 8 hours
Additional higher level: 6 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Gravitational potential energy
When two or more bodies interact, work is required to assemble them in their respective positions.
The total work done in bringing these bodies together is known as the gravitational potential energy of the system.
The gravitational potential energy of a system of two masses is defined as the work done in bringing the two masses from infinity to their respective positions.
In the given figure, as the bag moves upward, kinetic energy decreases while gravitational potential energy increases. At the highest point (B), kinetic energy becomes zero, and gravitational potential energy reaches its maximum.
Since the bag has no kinetic energy at point B, it begins to fall due to gravity. Initially, it falls slowly with low kinetic energy, but as it accelerates downward, kinetic energy increases.
At point A, the bag reaches its maximum speed, meaning kinetic energy is at its highest, while gravitational potential energy is nearly lost.
Upon impact with the ground, the kinetic energy transforms into heat and sound due to the collision.
It is important to note that as an object moves downward, its gravitational potential energy decreases while kinetic energy increases, and vice versa when moving upward.
Example:
(a) Find the escape velocity from a point 2.0 × 10⁷ m from the center of the Earth.
▶️ Answer/Explanation
Escape velocity formula:
\( v_e = \sqrt{\frac{2GM}{r}} \)
Substitute values:
\( v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{2.0 \times 10^7}} \)
\( v_e = \sqrt{3.985 \times 10^7} \approx \boxed{6313 \, \text{m/s}} \)
This is the minimum speed required to escape Earth’s gravity from this height without further propulsion.
(b) Work done in moving the satellite from orbit A (2.0 × 10⁷ m) to orbit B (3.0 × 10⁷ m)
▶️ Answer/Explanation
Use gravitational potential energy formula
\( E_{pA} = -\frac{GMm}{r_A} \), \( E_{pB} = -\frac{GMm}{r_B} \)
\( E_{pA} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200}{2.0 \times 10^7} \approx -2.39 \times 10^{10} \, \text{J} \)
\( E_{pB} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200}{3.0 \times 10^7} \approx -1.59 \times 10^{10} \, \text{J} \)
Work done = ΔE
\( W = E_{pB} – E_{pA} = (-1.59 \times 10^{10}) – (-2.39 \times 10^{10}) \)
\( W = \boxed{8.0 \times 10^9 \, \text{J}} \)
Interpretation:
This is the energy required to move the satellite from a lower to a higher orbit (gain in potential).
Gravitational potential energy
Think of potential energy as the capacity to do work. -And work is a force $F$ times a displacement $d$.
Recall the gravitational force from Newton:
If we multiply the above force by a distance $r$ we get
Note that EP is negative. Note also that EP = 0 when r = ∞.
EXAMPLE:
Find the gravitational potential energy stored in the Earth-Moon system.
▶️Answer/Explanation
SOLUTION:
Use $E_P=-\frac{G M m}{r}$.
$$
\begin{aligned}
E_P & =-\frac{\left(6.67 \times 10^{-11}\right)\left(5.98 \times 10^{24}\right)\left(7.36 \times 10^{22}\right)}{3.82 \times 10^8} \\
& =-7.68 \times 10^{28} \mathrm{~J} .
\end{aligned}
$$
Potential – gravitational
$
E_P=-\frac{G M m}{r} \text { where } G=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2} \quad \begin{gathered}
\text { gravitational } \\
\text { potential energy }
\end{gathered}
$
We now define gravitational potential as gravitational potential energy per unit mass:
This is why it is called “potential”.
FYI
The units of $\Delta V_g$ and $V_g$ are $\mathrm{J} \mathrm{kg}^{-1}$.
Gravitational potential is the work done per unit mass in moving a small mass from infinity to $r$. (Note that $V=0$ at $r=\infty$.)
Example:
Calculate the gravitational force between Earth (mass \( M = 5.97 \times 10^{24} \, \text{kg} \)) and a satellite (mass \( m = 600 \, \text{kg} \)) that is orbiting at an altitude of \( 300 \, \text{km} \) above Earth’s surface. Take the radius of Earth to be \( R = 6.37 \times 10^6 \, \text{m} \) and \( G = 6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \).
▶️ Answer/Explanation
Step 1: Total distance between centers of mass (r):
\( r = R + h = 6.37 \times 10^6 + 3.00 \times 10^5 = 6.67 \times 10^6 \, \text{m} \)
Step 2: Use Newton’s Law of Gravitation:
\( F = \frac{G M m}{r^2} \)
\( F = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(600)}{(6.67 \times 10^6)^2} \)
\( F \approx \boxed{5.37 \times 10^3 \, \text{N}} \)
Step 3: Gravitational field strength at that altitude:
\( g = \frac{GM}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.67 \times 10^6)^2} \approx \boxed{8.95 \, \text{N/kg}} \)
Step 4: Double-check force using \( F = mg \):
\( F = 600 \times 8.95 = \boxed{5.37 \times 10^3 \, \text{N}} \)
This confirms that the gravitational field strength and Newton’s Law yield consistent results.
Potential and potential energy – gravitational
FYI
A few words clarifying the gravitational potential energy and gravitational potential formulas are in order.
* Be aware of the difference in name. Both have “gravitational potential” in them and can be confused during problem solving.
* Be aware of the minus sign in both formulas.
* The minus sign is there so that as you separate two masses, or move farther out in space, their values increase (as in the last example).
* Both values are zero when r becomes infinitely large.
Example:
Which one of the following statements correctly defines the gravitational potential at a point P in a gravitational field?
A. The work done per unit mass in moving a small mass from point P to infinity.
B. The work done per unit mass in moving a small mass from infinity to point P.
C. The work done in moving a small mass from infinity to point P.
D. The work done in moving a small mass from point P to infinity.
▶️Answer/Explanation
Ans:
The correct answer is B.
- Gravitational potential at a point is defined as the work done per unit mass to bring a test mass from infinity (where the potential is zero) to that point in the gravitational field.