Gravitational Potential Energy and Gravitational Potential
Gravitational Potential Energy ($E_p$)
Gravitational potential energy
Gravitational potential energy is the energy stored due to the position of an object in a gravitational field.
Gravitational potential energy is always negative because it is defined relative to a point at infinity, where the potential energy is zero.
The formula for gravitational potential energy between two masses $m_1$ and $m_2$ separated by a distance $r$ is: $$E_p = -G \frac{m_1 m_2}{r}$$ where:
- $G$ is the gravitational constant ($6.67 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2}$).
- $r$ is the distance between the centers of the masses.
Consider a satellite of mass $1500 \, \text{kg}$ orbiting Earth at a distance of $7000 \, \text{km}$ from its center. Calculate its gravitational potential energy.
Solution
Substituting the given values:
$$E_p = -G \frac{M_{\text{Earth}} \cdot m_{\text{satellite}}}{r}$$
$$E_p = -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1500}{7.0 \times 10^6}$$
$$ \approx -8.57 \times 10^{10} \, \text{J}$$
Gravitational Potential ($V_g$)
Gravitational potential
Gravitational potential ($V_g$) is the work done per unit mass to bring a small test mass from infinity to a point in a gravitational field.
It is given by:
$$V_g = -G \frac{M}{r}$$
where:
- $M$ is the mass creating the gravitational field.
- $r$ is the distance from the center of the mass.
Gravitational potential is a scalar quantity and is measured in joules per kilogram (J kg⁻¹).
Calculate the gravitational potential at a distance of $10,000 \, \text{km}$ from Earth’s center.
Solution
$$V_g = -G \frac{M_{\text{Earth}}}{r} $$
$$= -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{1.0 \times 10^7} $$
$$\approx -4.0 \times 10^7 \, \text{J kg}^{-1}$$
Relationship Between $E_p$ and $V_g$
The gravitational potential energy $E_p$ of a mass $m$ at a point in a gravitational field is related to the gravitational potential $V_g$ at that point by:
$$E_p = m V_g$$
Example- A $200 \, \text{kg}$ satellite is at a point where the gravitational potential is $-5.0 \times 10^6 \, \text{J kg}^{-1}$.
- The gravitational potential energy is:
$$E_p = m V_g = 200 \times (-5.0 \times 10^6) = -1.0 \times 10^9 \, \text{J}$$
