Gravitational Potential Energy

This topic covers Gravitational Potential Energy.

Gravitational Potential

Gravitational Potential energy belongs to a pair of objects in a system (for instance a ball+ Earth system, galaxies of stars interacting gravitationally) and is equal to the work done against gravity. This potential energy is the energy associated within the particles inside a system and is not the same as rest or kinetic energies of the individual particles. Traditionally potential energy is represented by the symbol U and this page describes specific examples in which U is equal to the gravitational energy learned about in earlier pages to be approximately mg near the surface of the Earth or [math]\displaystyle{ F = - G \frac{m_1 m_2}{r^2}\ }[/math]
where:

• F is the force between the masses;
• G is the gravitational constant (6.674×10⁻¹¹ N · (m/kg)²);
• m₁ is the first mass;
• m₂ is the second mass;
• r is the distance between the centers of the masses.

The latter case is distance dependent and can be derived since force is the negative gradient of U. The negative indicates that potential energy decreases as particles get closer together

A Mathematical Model

In a system composed of two objects that are interacting with gravitational potential: [math]\displaystyle{ F = - G \frac{m_1 m_2}{r^2}\ }[/math]
where:

• F is the force between the masses;
• G is the gravitational constant (6.674×10⁻¹¹ N · (m/kg)²);
• m₁ is the first mass;
• m₂ is the second mass;
• r is the distance between the centers of the masses.

Close to the Surface of the Earth:

[math]\displaystyle{ U = mgh\! }[/math]

where U is the potential energy of the object assuming it is close to the surface of the Earth, m is the mass of the object, g is the acceleration of 9.8, and h is height.

Potential difference is derived to be:

[math]\displaystyle{ \,\Delta U = mg \Delta h.\ }[/math]

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

A ball of mass 100 grams is 7m above the ground, initially at rest( Ki=0) When the ball is 4 m above the ground what is the final Kinetic energy? Choose the ball+ Earth system. ==

The rest Energy is 0.

[math]\displaystyle{ 0=\Delta K_{(Earth)} + \Delta K_{(ball)} +\Delta U }[/math],

[math]\displaystyle{ 0= 0+(K_{(ball,f)} -0)+ \Delta(mgy) }[/math]

[math]\displaystyle{ 0= K_{(ball,f)} + (0.1 kg)*(9.8 N/M)*(-3 m) }[/math]

[math]\displaystyle{ U_g -2.94J }[/math]

This is used to solve for kinetic energy

[math]\displaystyle{ K_{(ball,f)} = 2.94 J }[/math]

Connectedness

1. http://voyager.jpl.nasa.gov/ Nasa uses gravitational energy to get probes to break through the heliosphere as in the voyager mission. Gravitational energy is very important to understand when sending probes to explore planets.
2. Biochem majors may find this information useful in understanding how chemical potential works.
3. Actual Application: http://www.forbes.com/sites/startswithabang/2015/11/02/move-over-hubble-gravity-itself-is-the-best-cosmic-telescope-of-all/

History

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See also

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