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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 negative when it is defined to be zero at infinity, since for two masses at a finite separation the potential energy is less than 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:
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
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:
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
$$V_g = -G \frac{M_{\text{Earth}}}{r} $$
$$= -\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{1.0 \times 10^7} $$
$$\approx -4.0 \times 10^7 \, \text{J kg}^{-1}$$
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:
$$E_p = m V_g$$
$$E_p = m V_g = 200 \times (-5.0 \times 10^6) = -1.0 \times 10^9 \, \text{J}$$
It is given by:
$$g = \frac{GM}{r^2}$$
Gravitational field strength is a vector quantity and is measured in newtons per kilogram (N kg⁻¹).
The negative sign indicates that the gravitational field points in the direction of decreasing potential.
If the gravitational potential decreases by $2.0 \times 10^6 \, \text{J kg}^{-1}$ over a distance of $500 \, \text{m}$, the field strength is:
$$g = -\frac{\Delta V_g}{\Delta r} = -\frac{-2.0 \times 10^6}{500} = 4000 \, \text{N kg}^{-1}$$
$$W = m \Delta V_g$$
where $\Delta V_g$ is the change in gravitational potential between the initial and final positions.
A $500 \, \text{kg}$ satellite is moved from a point where $V_g = -4.0 \times 10^6 \, \text{J kg}^{-1}$ to a point where $V_g = -2.0 \times 10^6 \, \text{J kg}^{-1}$.
The work done is:
$$W = m \Delta V_g$$
$$ = 500 \times \left((-2.0 \times 10^6) - (-4.0 \times 10^6)\right)$$
$$ = 1.0 \times 10^9 \, \text{J}$$
