Gravitational Energy

In near-Earth applications, potential energy induced by gravity can be calculated using $U = mgh$ . This, however, gets less accurate the further you get from Earth as gravity isn't constant.

Deriving $\Delta U$ and $U(r)$

Recall that work is defined as the integral of force with respect to displacement:

W_{AB} = \int_{A}^{B} \vec{F} \cdot d \vec{r}

Also recall that change in potential energy is equal to the negative of work done by a conservative force:

\Delta U = -W_{AB} = -\int_{A}^{B} \vec{F} \cdot d \vec{r}

Evaluating this integral from initial distance $r_1$ to final distance $r_2$ , we get:

\Delta U = GM_{E}m \left( \frac{1}{r_1} - \frac{1}{r_2} \right)

(1)

This formula for $\Delta U$ provides the change in potential energy of an object with mass $m$ moving from distance $r_1$ to distance $r_2$ relative to Earth's center.

To define absolute potential energy $U$ at a single position $r$ , rather than a difference between two points, we must establish a reference point where potential energy is zero. We can define $U = 0$ at an infinite distance away ( $r \to \infty$ ).

Expanding $\Delta U = U_f - U_i$ where the initial position is $r_1 = r$ and the final position is sent out to infinity ( $r_2 = \infty$ , so $U_f = 0$ ):

U_f - U_i = GM_{E}m \left( \frac{1}{r_1} - \frac{1}{r_2} \right)

0 - U = GM_{E}m \left( \frac{1}{r} - \frac{1}{\infty} \right)

-U = GM_{E}m \left( \frac{1}{r} - 0 \right)

U = -\frac{GM_{E}m}{r}

(2)

Notice that in Equation 2, potential energy is negative. This is a direct consequence of choosing $U = 0$ at $r \to \infty$ .

Because gravity is an attractive force, it does positive work as an object falls inward toward Earth. Since a loss in potential energy corresponds to positive work ( $\Delta U = -W$ ), moving closer to Earth decreases potential energy further below zero.

An object at a finite distance $r$ is bound by Earth's gravitational field. It would require an input of positive energy equal to $\frac{GM_E m}{r}$ to pull it completely free to infinity.

Example 1

A $\text{3000 kg}$ mass is boosted from the surface of Earth to a height of $\text{500 km}$ . How much energy was expended?

Solution Click to unblur

We must first recognize this problem is asking for $\Delta U$ , not an absolute $U$ . This is because the mass starts on the surface, therefore $U_i \neq 0$ . We say $U_i$ is the potential energy of the mass on the surface of Earth, and $U_f$ is the potential energy at a height of $\text{500 km}$ .

\Delta U = GM_{E}m \left( \frac{1}{r_E} - \frac{1}{r_E + r} \right)

Plugging in our known values for $G$ , $M_E$ , $m$ , $r_E$ , and $r$ :

\Delta U = \left( 6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2} \right) (5.97 \cdot 10^{24} kg) (3000kg) \left( \frac{1}{6.37 \cdot 10^6 m} - \frac{1}{6.37 \cdot 10^6 m + 5 \cdot 10^5m} \right)

\Delta U \approx 1.36 \cdot 10^{10} J

Conservation of Energy

Because gravity is a conservative force, we can apply the conservation of energy laws to gravity at extended distances.

Recall that the total mechanical energy for an object is the sum of its kinetic and potential energies:

E_{mech} = K + U

Also recall that for conservation of energy, the initial total mechanical energy must equal the final total mechanical energy:

K_i + U_i = K_f + U_f

Substituting in the kinetic energy for a mass in motion, as well as our definition of $U$ from Equation 2:

\frac{1}{2} m v_i^2 - \frac{GMm}{r_i} = \frac{1}{2}mv_f^2 - \frac{GMm}{r_f}

(3)

Note that $M_E$ from Equation 2 was replaced with a generic $M$ , as the equation doesn't strictly apply to Earth. It is still assumed that $m<M$ .

Bound vs. Unbound Systems

An object's trajectory is determined by its total mechanical energy $E = K + U$ :

If $E < 0$ , the object does not have enough kinetic energy to break away from the gravitational potential energy. It will remain in a closed orbit.
If $E = 0$ , the object has just enough kinetic energy to reach $r \to \infty$ where it comes to a complete stop ( $v_f = 0$ ). This represents the threshold for escape.
If $E > 0$ , the object escapes to infinity and still retains leftover kinetic energy ( $v_f > 0$ ), following a hyperbolic trajectory.

Escape Velocity

The escape velocity ( $v_{esc}$ ) is the minimum speed an object must be given at initial radius $r$ to escape a central mass $M$ completely without further propulsion.

This occurs precisely at the boundary condition where total mechanical energy is zero ( $E = 0$ ).

Deriving Escape Velocity

Setting total energy $E = 0$ at distance $r$ :

K + U = 0

\frac{1}{2}m v_{esc}^2 - \frac{GMm}{r} = 0

Solving for $v_{esc}$ :

\frac{1}{2}m v_{esc}^2 = \frac{GMm}{r}

v_{esc}^2 = \frac{2GM}{r}

v_{esc} = \sqrt{\frac{2GM}{r}}

(3)

Example 2

Calculate the escape velocity from the surface of the Moon given $M_M = 7.35 \cdot 10^{22} \text{ kg}$ and $r_M = 1.74 \cdot 10^{6} \text{ m}$ .

Solution Click to unblur

Using Equation 3:

v_{esc} = \sqrt{\frac{2GM_M}{r_M}} = \sqrt{\frac{2 \left(6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}\right) \left(7.35 \cdot 10^{22} \text{ kg}\right)}{1.74 \cdot 10^{6}\text{ m}}} \approx 2.38 \cdot 10^3 \text{ m/s}

Example 3

A rocket is launched vertically from Earth's surface with an initial speed $v_i = 8000 \text{ m/s}$ . Ignoring air resistance, what maximum height above Earth's surface does it reach?

Hint Click to unblur

Use conservation of energy. Remember that when a rocket reaches its maximum height (apex), $v_f = 0$ .

Solution Click to unblur

Start with energy conservation:

\frac{1}{2} m v_i^2 - \frac{GM_E m}{r_i} = \frac{1}{2}mv_f^2 - \frac{GM_E m}{r_f}

Cancel out mass $m$ :

\frac{1}{2} v_i^2 - \frac{GM_E}{r_i} = \frac{1}{2}v_f^2 - \frac{GM_E}{r_f}

Since $v_f = 0$ at the apex, omit the final kinetic energy term:

\frac{1}{2} v_i^2 - \frac{GM_E}{r_i} = - \frac{GM_E}{r_f}

Rearrange to isolate $\frac{1}{r_f}$ :

\frac{GM_E}{r_i} - \frac{1}{2} v_i^2 = \frac{GM_E}{r_f}

\frac{1}{r_i} - \frac{v_i^2}{2GM_E} = \frac{1}{r_f}

Combine terms on the left:

\frac{2GM_E - r_i v_i^2}{2 r_i GM_E} = \frac{1}{r_f}

Invert to solve for $r_f$ :

r_f = \frac{2 r_i GM_E}{2 GM_E - r_i v_i^2}

Plugging in known values ( $r_i = 6.371 \cdot 10^6 \text{ m}$ , $G = 6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}$ , $M_E = 5.972 \cdot 10^{24} \text{ kg}$ , $v_i = 8000 \text{ m/s}$ ):

r_f = \frac{2 (6.371 \cdot 10^6 \text{ m})(6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2})(5.972 \cdot 10^{24} \text{ kg})}{2 (6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2})(5.972 \cdot 10^{24} \text{ kg}) - (6.371 \cdot 10^6 \text{ m})(8000 \text{ m/s})^2}

r_f \approx 1.304 \cdot 10^7 \text{ m}

Finally, calculate the height $h$ above Earth's surface:

h = r_f - r_i = 1.304 \cdot 10^7 \text{ m} - 6.371 \cdot 10^6 \text{ m} \approx 6.67 \cdot 10^6 \text{ m}

Example 4

A $1000 \text{ kg}$ asteroid far outside the solar system approaches Earth from deep space with an initial speed of $v_\infty = 3000 \text{ m/s}$ . How fast will it be moving right before it impacts Earth's surface ( $r \approx R_E$ )?

Hint Click to unblur

Because the asteroid originates at "deep space" ( $r \to \infty$ ), its initial potential energy $U_i = 0$ . Since total energy $E > 0$ , it follows an unbound hyperbolic trajectory.

Solution Click to unblur

Apply conservation of mechanical energy between deep space ( $r_i \to \infty$ ) and Earth's surface ( $r_f = R_E$ ):

K_i + U_i = K_f + U_f

\frac{1}{2}m v_\infty^2 - 0 = \frac{1}{2}m v_f^2 - \frac{GM_E m}{R_E}

Cancel out mass $m$ from all terms:

\frac{1}{2} v_\infty^2 = \frac{1}{2} v_f^2 - \frac{GM_E}{R_E}

Multiply by 2 and solve for final velocity $v_f$ :

v_f^2 = v_\infty^2 + \frac{2GM_E}{R_E}

v_f = \sqrt{v_\infty^2 + \frac{2GM_E}{R_E}}

Note that $\frac{2GM_E}{R_E}$ is equal to $v_{esc}^2$ , showing that impact velocity for an unbound object is always $v_f = \sqrt{v_\infty^2 + v_{esc}^2}$ .

Plugging in known values ( $v_\infty = 3000 \text{ m/s}$ , $G = 6.674 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}$ , $M_E = 5.972 \cdot 10^{24} \text{ kg}$ , $R_E = 6.371 \cdot 10^6 \text{ m}$ ):

v_f = \sqrt{(3000 \text{ m/s})^2 + \frac{2 (6.674 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2})(5.972 \cdot 10^{24} \text{ kg})}{6.371 \cdot 10^6 \text{ m}}}

v_f = \sqrt{9 \cdot 10^6 \text{ m}^2/\text{s}^2 + 1.251 \cdot 10^8 \text{ m}^2/\text{s}^2} \approx 1.16 \cdot 10^4 \text{ m/s}

The asteroid impacts at approximately $11.6 \text{ km/s}$ (or $11,580 \text{ m/s}$ ).

Example 5

A $1500 \text{ kg}$ space probe is launched from Earth's surface on a parabolic trajectory with total mechanical energy $E = 0$ .

a.) What launch velocity $v_i$ is required?

b.) What will its speed be when it is very far from Earth ( $r \to \infty$ )?

Hint Click to unblur

A total mechanical energy of $E = 0$ represents the exact boundary between bound ( $E < 0$ ) and unbound ( $E > 0$ ) states. This corresponds directly to the escape velocity equation.

Solution A Click to unblur

Set the total initial energy to zero:

E = K_i + U_i = 0

\frac{1}{2}m v_i^2 - \frac{GM_E m}{R_E} = 0

Cancel mass $m$ and solve for $v_i$ :

\frac{1}{2} v_i^2 = \frac{GM_E}{R_E}

v_i = \sqrt{\frac{2GM_E}{R_E}}

Substitute known values ( $G = 6.674 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}$ , $M_E = 5.972 \cdot 10^{24} \text{ kg}$ , $R_E = 6.371 \cdot 10^6 \text{ m}$ ):

v_i = \sqrt{\frac{2(6.674 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2})(5.972 \cdot 10^{24} \text{ kg})}{6.371 \cdot 10^6 \text{ m}}}

v_i \approx 1.12 \cdot 10^4 \text{ m/s} \quad (11.2 \text{ km/s})

Solution B Click to unblur

As $r \to \infty$ , potential energy approaches zero ( $U_\infty = 0$ ).

By energy conservation:

E = K_\infty + U_\infty = 0

\frac{1}{2}m v_\infty^2 + 0 = 0 \implies v_\infty = 0 \text{ m/s}

The probe comes to a rest relative to Earth as it approaches infinite distance.

Gravitational Energy

Deriving ΔU\Delta UΔU and U(r)U(r)U(r)

Example 1

Conservation of Energy

Bound vs. Unbound Systems

Escape Velocity

Deriving Escape Velocity

Example 2

Example 3

Example 4

Example 5

Deriving $\Delta U$ and $U(r)$