Gravitational Potential Energy

Potential energy is the energy stored in an object because of its position. For example, if you lift a ball from the ground, it has potential energy because it can fall back down.

Near the Earth's surface, we calculate potential energy using the formula:

PE = mgh

where:

• m = mass of the object
• g = acceleration due to gravity (9.8 m/s²)
• $h$ = height above the ground

However, when an object is at a very high altitude (far from Earth), gravity weakens. In such cases, the formula for potential energy is more accurate as:

PE =  \[-\frac{\mathrm{GMm}}{\mathrm{R+h}}\]

where:

• G = gravitational constant
• M = mass of the Earth
• R = radius of the Earth
• h = height from Earth's surface

At an infinite distance from the Earth, the gravitational force becomes negligible, and the potential energy is assumed to be zero. As a result, potential energy at any finite distance is negative, indicating that work must be done to move the object to infinity.

When an object is thrown upwards, its velocity decreases due to Earth's gravitational attraction. If the initial velocity is not sufficient, the object eventually falls back. The maximum height reached depends on the initial velocity.

If the initial velocity is continuously increased, the object reaches greater heights. Beyond a certain velocity, the object overcomes Earth's gravitational pull completely and never returns. This velocity is termed escape velocity.

Using the principle of conservation of energy: On Earth's surface,

• Kinetic energy: `(1)/(2)`\[mv_{esc}^2\]
• Potential energy = -\[\frac{\mathrm{GMm}}{\mathrm{R}}\]
• Total energy = E₁ = Kinetic energy + Potential energy

`(1)/(2)`\[mv_{esc}^2\] - \[\frac{\mathrm{GMm}}{\mathrm{R}}\]

At infinite distance:

• Kinetic energy: 0
• Potential energy: 0
• Total energy: E²=0

Applying conservation of energy: E₁= E₂

`(1)/(2)`\[mv_{esc}^2\] - \[\frac{\mathrm{GMm}}{\mathrm{R}}\] = 0

Solving for ν_esc​:

\[\mathrm{v}_{\mathrm{esc}}^{2}=\quad\frac{2\mathrm{~GM}}{\mathrm{R}}\]

\[\mathbf{v}_{\mathrm{esc}}=\sqrt{\frac{2\mathrm{GM}}{\mathrm{R}}}\]

\[=\sqrt{2\mathrm{g}\mathrm{R}}\]

Substituting values for Earth:

ν_esc \[=\sqrt{(2\mathrm{x}9.8\mathrm{x}6.4\mathrm{x}10^{6})}\] = 11.2 km/s

Thus, an object must travel at 11.2 km/s or more to escape Earth's gravitational field. Spacecraft and satellites require an initial velocity greater than the escape velocity to leave Earth and reach space or other celestial bodies.