Potential energy of gravitational force

In summary, the conversation discusses a problem involving a bead sliding along the x-axis between two spheres of mass M, which are equidistant from the axis at a distance of d. The potential energy of the bead is found by integrating the force of gravity and can be expressed as a function of the distance from the bead to one of the spheres (r) and the radius of the spheres (R_M). The potential energy must also be multiplied by 2 to account for the x and y components of the gravitational forces. To find the speed at the origin, the potential energy must be set in terms of d. The conversation also mentions that the y component of the gravitational forces cancel out and the potential for the x component must be doubled.
  • #1
diablo2121
10
0

Homework Statement


A bead of mass m slides along the x-axis between two spheres of mass M equidistant from the x-axis (distance d) and attract the bead gravitationally.

a. Find the potential energy of the bead.

b. The bead is released at x = 3d with an initial velocity toward the origin. Find the speed at the origin.

c. Find the frequency of small oscillations of the bead about the origin.

Homework Equations


[tex]U = -\int F dx[/tex]

[tex]F_{grav} = \frac{-GMm}{r^2}[/tex]

[tex]E = U + K[/tex]

[tex]K = \frac{1}{2}mv^2[/tex]

The Attempt at a Solution


Integrating the force of gravity, I find that the potential energy of the force is [tex]\frac{GMm}{r^2}*{R_{M}}^2*(\frac{1}{R_{M}} - \frac{1}{r})[/tex] where [tex]r[/tex] is the distance straight from [tex]m[/tex] to one of the spheres [tex]M[/tex] and [tex]R_M[/tex] is the radius of [tex]M[/tex]. This is fine, but I'm stuck on how to set the potential in terms of d, which I will need to find the velocity in the next part.

Conceptually, I know the y component of the gravitational forces cancel out since the spheres are equal mass and distance away. Also, the potential I get for the x component will need to be doubled since there are two masses.
 
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  • #2
Wow, once you learn Latex, it's a lot easier to format your equations. I hope that helps the equations a little better to understand.
 
  • #3
Isn't the r distance really given by

[tex] r = \sqrt{x^2 + \frac{d^2}{4}} [/tex]
 

FAQ: Potential energy of gravitational force

What is potential energy of gravitational force?

The potential energy of gravitational force is the energy that an object possesses due to its position in a gravitational field. It is a measure of the work that would be required to move the object from its current position to a reference point, typically infinity, where the gravitational force is zero.

How is potential energy of gravitational force calculated?

The potential energy of gravitational force is calculated using the formula PE = mgh, where PE is potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

What factors affect the potential energy of gravitational force?

The potential energy of gravitational force is affected by the mass of the object, the gravitational constant, and the distance between the object and the reference point. The higher the mass and the greater the distance, the higher the potential energy will be.

How does potential energy of gravitational force relate to kinetic energy?

The potential energy of gravitational force and kinetic energy are two forms of energy that are related through the Law of Conservation of Energy. When an object is in motion, its potential energy is converted into kinetic energy. Similarly, when an object is at rest, its kinetic energy is converted into potential energy.

What are some real-world applications of potential energy of gravitational force?

The potential energy of gravitational force is utilized in many engineering and scientific applications such as hydroelectric power plants, roller coasters, and space missions. It is also used in everyday activities such as lifting objects, climbing stairs, and throwing a ball.

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