Why is gravitational potential energy always taken as negative?

Gravitational force is purely attractive. By convention, the reference point of zero potential energy is chosen at an infinite separation (r = ∞, U = 0). As two masses are brought closer from infinity, the attractive gravitational field does positive work, which means the potential energy of the system decreases below zero, resulting in a negative potential energy: U(r) = -G M m / r. The negative sign signifies that the system is bound.

Does escape velocity depend on the angle of projection or the mass of the projectile?

No. The escape speed v_e = √(2GM/R) = √(2gR) depends solely on the mass and radius of the celestial body from which the launch occurs. It is completely independent of: 1. The mass of the projected body; 2. The direction or angle of projection (whether fired vertically, horizontally, or at 45°).

§ 7.3NCERT Class 11 · Physics · Chapter 7

Gravitational Potential Energy, Gravitational Potential & Escape Speed

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 8 min read

In gravitational fields, understanding work, potential energy, and escape velocity is essential for analyzing planetary systems, spaceflight trajectories, and projectile motion. Because gravity is an attractive conservative central force, gravitational potential energy is defined relative to an arbitrary reference at infinity.

1. Gravitational Potential Energy (U)

The gravitational potential energy of a system of two masses M and m separated by a distance r is defined as the work done by an external agent in bringing mass m from infinity to distance r slowly (without acceleration) in the gravitational field of mass M:

U(r) = −G M m / r (with reference U(∞) = 0)

Key observations regarding potential energy:

The negative sign indicates a bound system: positive work must be done against the gravitational pull to separate the masses to infinity.
If r increases, U becomes less negative (i.e., potential energy increases).
If r decreases, U becomes more negative (potential energy decreases).

Work Done in Lifting a Mass to Height h

When a body of mass m is lifted from Earth's surface (r₁ = R) to an altitude h (r₂ = R + h), the change in potential energy is:

ΔU = U(R + h) − U(R) = [−G M m / (R + h)] − [−G M m / R]

ΔU = G M m [1/R − 1/(R + h)] = G M m h / [R (R + h)]

General Formula: ΔU = m g h / (1 + h/R)
When h ≪ R (near Earth's surface): 1 + h/R ≈ 1 ⇒ ΔU ≈ m g h (familiar introductory physics formula).

Gravitational Potential Energy Well & Escape Mechanics

Potential well U(r) and energy conservation during escape from a celestial body

GRAVITATIONAL POTENTIAL WELL:
----------------------------
U = 0  --------------------------------------------- [At r = ∞ (Free)]
               ^
               |   ΔU = m g h / (1 + h/R)
               |
U(R+h) --------+-------------------- [At Height h]
               |
               |
U(R)   --------+---------- [At Surface r = R: U = -GMm/R]
               |
               v   (Deep Bound State)

ESCAPE SPEED ENERGY CONSERVATION:
--------------------------------
Total Energy at Surface = Total Energy at Infinity
(1/2) m ve² + U(R) = 0 + 0  -->  (1/2) m ve² - G M m / R = 0
==>  ve = √(2 G M / R) = √(2 g R)

Earth: ve ≈ 11.2 km/s | Moon: ve ≈ 2.38 km/s

Atmospheric Retention Law: A planet retains an atmosphere only if the thermal root-mean-square speed of its atmospheric gas molecules is significantly less than the escape velocity (v_rms < v_e / 5). On the Moon, v_rms of lighter gases exceeds 2.38 km/s, allowing all atmospheric gases to escape into space.

2. Gravitational Potential (V)

The gravitational potential V at any point in a gravitational field is the potential energy per unit mass placed at that point, or the work done in bringing a unit test mass from infinity to that point:

V(r) = U(r) / m = −G M / r (for r ≥ R)

Relation between gravitational field I and potential V:

I = −dV / dr or V(r) = −∫ I dr

Gravitational Potential for a Uniform Solid Sphere of Mass M and Radius R

Region	Position Condition	Gravitational Potential Formula
Outside Sphere	r ≥ R	V(r) = −G M / r
At Surface	r = R	V_s = −G M / R
Inside Sphere	r < R	V(r) = −[G M / (2R³)] (3R² − r²)
At Center of Sphere	r = 0	V_c = −3/2 (G M / R) = 1.5 V_s

3. Escape Speed (ve)

The escape speed is defined as the minimum initial speed with which a body must be projected from the surface of a planet so that it just overcomes the gravitational attraction and escapes to infinity.

Derivation via Mechanical Energy Conservation

Let a body of mass m be projected with speed v_e from the surface of Earth (mass M, radius R):

Initial mechanical energy at surface: E_i = K_i + U_i = ½ m v_e² − G M m / R
Final mechanical energy at infinity (just reaching infinity with zero residual speed): E_f = 0 + 0 = 0

By conservation of total mechanical energy (E_i = E_f):

½ m v_e² − G M m / R = 0 ⇒ ½ m v_e² = G M m / R

v_e = √(2 G M / R) = √(2 g R)

Fundamental Characteristics of Escape Speed

Independent of Projectile Mass: A grain of sand and a massive rocket have the identical escape speed from Earth: v_e ≈ 11.2 km/s.
Independent of Projection Angle: Since energy is a scalar quantity, whether projected vertically, at 30°, 45°, or 60° to the horizontal, the escape speed remains √(2gR) (neglecting atmospheric air drag and Earth's rotation).
In terms of planetary density ρ: v_e = √[2 G (4/3 π R³ ρ) / R] = R √(8/3 π G ρ) ∝ R √ρ.

4. Solved High-Yield JEE / NEET Practice Questions

Question 1: The work done in lifting a mass m from the surface of the Earth to a height equal to the radius of the Earth (h = R) is:

(A) m g R

(B) (1/2) m g R

(D) (1/4) m g R

Show answer

Correct Answer: (B) (1/2) m g R

Step-by-step Solution:
1. Apply the exact formula for change in gravitational potential energy:
ΔU = m g h / (1 + h/R).
2. Substitute h = R into the expression:
ΔU = m g R / (1 + R/R) = m g R / (1 + 1) = (1/2) m g R.
Note: Using the naive introductory formula ΔU = mgh = mgR gives an error of 100% because g is not constant over such large distances!

Question 2: If the acceleration due to gravity on a planet is twice that on Earth and its radius is half that of Earth, the escape speed from the planet will be:

(A) 2 ve

(B) ve

(D) √2 ve

Show answer

Correct Answer: (B) ve

Step-by-step Solution:
1. Recall the escape speed formula: v_e = √(2 g R).
2. For the new planet: g' = 2g and R' = R/2.
3. Substitute the values into the formula:
v'_e = √[2 g' R'] = √[2 (2g) (R/2)] = √[2 g R] = ve.
The escape speed remains completely unchanged!

Question 3: A body is thrown with a velocity equal to 3 times the escape velocity from the surface of Earth (v = 3 ve). What will be its residual velocity in interstellar space (at infinity)?

(A) 2 ve

(B) 2√2 ve

(D) √10 ve

Show answer

Correct Answer: (B) 2√2 ve

Step-by-step Solution:
1. Apply conservation of mechanical energy between Earth's surface and infinity:
½ m v² − G M m / R = ½ m v_∞² + 0.
2. Recognize that G M m / R = ½ m v_e²:
½ m v² − ½ m v_e² = ½ m v_∞² ⇒ v_∞² = v² − v_e².
3. Given v = 3 v_e:
v_∞² = (3 v_e)² − v_e² = 9 v_e² − v_e² = 8 v_e².
4. Taking the square root:
v_∞ = √8 v_e = 2√2 ve.

5. Frequently Asked Questions (FAQs)

1. What is the value of gravitational potential at the center of the Earth?

At the center of a uniform solid sphere (Earth), V_c = −1.5 (G M / R) = 1.5 V_s. That is, the potential at the center is 1.5 times the potential at the surface.

2. If a body is projected at 45° to the horizontal with speed ve, will it escape?

Yes! The escape velocity derivation relies purely on scalar mechanical energy conservation. Any trajectory with speed ≥ √(2gR) has non-negative total mechanical energy and will escape to infinity, regardless of launch angle (assuming no obstacle or atmospheric resistance).

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