Gravitational Potential Energy: Videos & Practice Problems

Gravitational potential energy is defined by the equation $U=-GmM/r$, where G is the universal gravitational constant, m and M are the masses, and r is the distance between their centers. This equation is crucial for solving problems involving changing energies and velocities, especially when gravitational forces vary with distance. Energy conservation principles apply, allowing the calculation of final velocities during gravitational interactions, such as an asteroid falling to Earth.

Gravitational Potential Energy

Gravitational potential energy is a crucial concept in understanding how masses interact through gravitational forces. When analyzing problems involving gravitational potential energy, it is essential to use the correct equation, especially when multiple masses and distances are involved. The new equation for gravitational potential energy is given by:

U = -\frac{G m_1 m_2}{r}

In this equation, G represents the gravitational constant, m₁ and m₂ are the masses of the two objects, and r is the distance between their centers of mass. The negative sign indicates that gravitational potential energy decreases as the distance between the masses decreases. It is important to note that r is defined as the distance from the center of mass, which can be expressed as R + h when considering a planet, where R is the radius of the planet and h is the height above the surface.

For small changes in height, the gravitational potential energy can be approximated using the simpler formula:

U = mgh

where g is the acceleration due to gravity, typically approximated as 9.8 m/s² near the Earth's surface. However, this approximation is only valid when the height changes are minimal, and the gravitational force can be considered constant.

When solving problems involving the motion of objects under the influence of gravity, such as an asteroid falling towards Earth, it is crucial to apply the principle of conservation of energy. In such scenarios, the total mechanical energy (kinetic plus potential energy) remains constant, provided no non-conservative forces (like friction) are doing work. The initial potential energy of the system can be expressed as:

U_i = -\frac{G m_{Earth} m_{asteroid}}{r_{initial}}

and the final kinetic energy when the asteroid impacts the Earth can be expressed as:

K_f = \frac{1}{2} m_{asteroid} v_{final}^2

By setting the initial potential energy equal to the final kinetic energy, we can derive the final velocity of the asteroid just before impact. The equation can be rearranged to isolate v_final:

\frac{G m_{Earth} m_{asteroid}}{R} - \frac{G m_{Earth} m_{asteroid}}{r_{initial}} = \frac{1}{2} m_{asteroid} v_{final}^2

After simplifying and canceling out the mass of the asteroid, we can solve for v_final:

v_{final} = \sqrt{2G m_{Earth} \left(\frac{1}{R} - \frac{1}{r_{initial}}\right)}

In this equation, R is the radius of the Earth, and r_initial is the initial distance from the center of the Earth to the asteroid. By substituting the known values for G, m_Earth, R, and r_initial, we can calculate the final velocity of the asteroid as it impacts the Earth.

In summary, understanding gravitational potential energy and its application in energy conservation principles allows for the effective analysis of problems involving gravitational interactions between masses. This approach is essential for accurately determining the velocities and energies of objects in motion under gravitational influence.

How much energy is required to move a 1000-kg object from Earth's surface to a height twice Earth's radius?

• A. $3.13\times 10^{10}$ J
• B. $4.17\times 10^{10}$ J
• C. $7.83\times 10^{23}$ J
• D. $8.46\times 10^{23}$ J

Answer: (B)

Collision of Two Small Planets

In this scenario, we are analyzing the gravitational interaction between two identical masses, each with a mass of \( m = 7 \times 10^{22} \) kg, initially separated by a distance of \( r_i = 5 \times 10^{10} \) m. As they drift towards each other due to gravitational attraction, we aim to determine their final speed \( v_f \) at the moment of collision.

Since the forces involved are variable, we will utilize the principle of conservation of energy rather than kinematics. The total mechanical energy of the system, which includes both kinetic and potential energy, remains constant in the absence of non-conservative forces. Initially, both masses are at rest, so their initial kinetic energy \( KE_i \) is zero:

$$ KE_i = 0 $$

The initial gravitational potential energy \( PE_i \) between the two masses can be expressed as:

$$ PE_i = -\frac{G m^2}{r_i} $$

where \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) is the gravitational constant.

As the masses collide, they will have a final kinetic energy \( KE_f \) and a final gravitational potential energy \( PE_f \). The final kinetic energy for both masses can be expressed as:

$$ KE_f = \frac{1}{2} m v_f^2 + \frac{1}{2} m v_f^2 = m v_f^2 $$

At the point of collision, the distance between the centers of the two masses is equal to \( 2R \), where \( R \) is the radius of each mass. Thus, the final potential energy is:

$$ PE_f = -\frac{G m^2}{2R} $$

Applying the conservation of energy principle, we set the initial total energy equal to the final total energy:

$$ KE_i + PE_i = KE_f + PE_f $$

Substituting the expressions for kinetic and potential energy, we have:

$$ 0 - \frac{G m^2}{r_i} = m v_f^2 - \frac{G m^2}{2R} $$

Rearranging this equation to isolate \( v_f^2 \), we get:

$$ \frac{G m^2}{r_i} - \frac{G m^2}{2R} = m v_f^2 $$

Dividing through by \( m \) and simplifying gives:

$$ v_f^2 = \frac{G m}{r_i} - \frac{G m}{2R} $$

Factoring out \( Gm \) leads to:

$$ v_f^2 = G m \left( \frac{1}{r_i} - \frac{1}{2R} \right) $$

Finally, taking the square root provides the final speed:

$$ v_f = \sqrt{G m \left( \frac{1}{r_i} - \frac{1}{2R} \right)} $$

Substituting the known values, we find:

$$ v_f = \sqrt{6.67 \times 10^{-11} \times 7 \times 10^{22} \left( \frac{1}{5 \times 10^{10}} - \frac{1}{4 \times 10^{6}} \right)} $$

After performing the calculations, the final velocity of each planet as they collide is:

$$ v_f \approx 1.08 \times 10^{3} \, \text{m/s} $$

This result indicates the speed at which both planets collide due to their mutual gravitational attraction.

You launch a rocket with an initial speed of $5\times 10^3$ m/s from Earth's surface. At what height above the Earth will it have ¼ of its initial launch speed? Assume the rocket's engines shut off after launch.

• A. $9.57\times 10^5$ m
• B. $1.47\times 10^6$ m
• C. $7.84\times 10^6$ m
• D. $2.02\times 10^{22}$ m

Answer: (B)

Here’s what students ask on this topic:

What is the formula for gravitational potential energy?

The formula for gravitational potential energy (U) is given by:

$U=-GmM/r$

where:

• $G$ is the universal gravitational constant (6.67430 × 10^-11 m³ kg^-1 s^-2).
• $m$ and $M$ are the masses of the two objects.
• $r$ is the distance between the centers of the two masses.

How does gravitational potential energy change with distance?

Gravitational potential energy (U) is inversely proportional to the distance (r) between two masses. As the distance increases, the magnitude of the gravitational potential energy decreases (becomes less negative), and as the distance decreases, the magnitude of the gravitational potential energy increases (becomes more negative). This relationship is described by the formula:

$U=-GmM/r$

Why can't we use kinematics to solve problems involving gravitational potential energy?

Kinematics assumes constant acceleration, which is not valid for gravitational interactions over large distances. In gravitational problems, the force and acceleration vary with distance according to Newton's law of gravitation. Therefore, we use energy conservation principles instead, applying the gravitational potential energy formula:

$U=-GmM/r$

This allows us to account for the changing gravitational force and solve for variables like final velocity.

How do you calculate the impact speed of an asteroid falling to Earth using gravitational potential energy?

To calculate the impact speed of an asteroid falling to Earth, use energy conservation principles. Set the initial potential energy and final kinetic energy equal:

$\frac{-GmM}{r}=\frac{1}{2}m{v}^2$

Rearrange to solve for the final velocity (v):

$v=\sqrt{\frac{2GM(\frac{1}{r}-\frac{1}{r}}{1}}$

where $r$ is the Earth's radius and $r_i$ is the initial distance from the Earth's center.

What is the significance of the negative sign in the gravitational potential energy formula?

The negative sign in the gravitational potential energy formula:

$U=-GmM/r$

indicates that gravitational potential energy is always negative, reflecting the attractive nature of gravity. It signifies that work must be done to separate the masses to an infinite distance, where the potential energy would be zero. The more negative the value, the stronger the gravitational attraction between the masses.