Derivation of gravitational potential energy

Actually andrew sir there is a problem which i want to discuss with you sir because one of my collegue said to me that gravity of earth is directly proportional to the radius he further says to me that he prove it by his geophysics fellow but my thinking is that gravity is inversly proportional to the radius kindly sir help me from this issue or give me a valid example i am very thankful to you if you help me. G is the universal constant for the gravitational force. It never changes. The units for G are m^3/(kg*s^2) g is the local acceleration due to gravity between 2 objects. The unit for g is m/s^2 an acceleration. The 9.8 m/s^2 is the acceleration of an object due to gravity at sea level on earth. You get this value from the Law of Universal Gravitation. Force = m*a = G(M*m)/r^2 Here you use the radius of the earth for r, the distance to sea level from the center of the earth, and M is the mass of the earth. Notice that little m cancels out on both sides of the equation. m*a=G(M*m)/r^2 a=G*M/r^2 If you put in the mass of the earth and the radius to sea level you will get 9.8 m/s^2 for a. This is what we call little g. Notice if you change your radius that the acceleration(g) will fall off as 1/r^2. If you are twice as far out 2*r, you will have 4 times less gravitational acceleration. That is why g can change from place to place on earth. If you are on Mt. Everest your radius will be larger than if you were in Death Valley.. for example. If you want to know how G came a...

Learning Objectives By the end of this section, you will be able to: • Determine changes in gravitational potential energy over great distances • Apply conservation of energy to determine escape velocity • Determine whether astronomical bodies are gravitationally bound We studied gravitational potential energy in g remained constant. We now develop an expression that works over distances such that g is not constant. This is necessary to correctly calculate the energy needed to place satellites in orbit or to send them on missions in space. Gravitational Potential Energy beyond Earth We defined work and potential energy in Δ U = m g ( y 2 − y 1 ) Δ U = m g ( y 2 − y 1 ). This works very well if g does not change significantly between y 1 y 1 and y 2 y 2. We return to the definition of work and potential energy to derive an expression that is correct over larger distances. Recall that work ( W) is the integral of the dot product between force and distance. Essentially, it is the product of the component of a force along a displacement times that displacement. We define Δ U Δ U as the negative of the work done by the force we associate with the potential energy. For clarity, we derive an expression for moving a mass m from distance r 1 r 1 from the center of Earth to distance r 2 r 2. However, the result can easily be generalized to any two objects changing their separation from one value to another. Consider m from a distance r 1 r 1 from Earth’s center to a distance that is...

This post focuses on the Derivation of Gravitational potential energy equations when the reference varies from the planet’s surface to infinity. Gravitational potential energy, E p, is the energy of a mass due to its position within a gravitational field. This energy can be released (and converted into kinetic energy) when the body is allowed to fall. In this post, let’s find out the concepts & equations of Gravitational potential energy when the reference point varies from the planet’s surface to infinity. We will discuss & derive 2 equations: (1) E p = mgh (2) E p = – G m 1m 2 / r Gravitational potential energy E p at some point x above the ground = work done to move to the point x with the vertical height h from the ground = force required × distance moved (since work W = Fr) Gravitational potential energy E p = (mg) × h = mgh fig 1: Gravitational potential energy Hence, in this case, E p = mgh. Here, We chose the ground as our starting point because this is our defined zero level; that is, the place where E p = 0. Note that since work must be done on the object to lift it, it acquires energy. Hence, at point x, E p is greater than zero. See also The formula for acceleration due to gravity at height h - with derivation On a larger, planetary-scale we need to rethink our approach. Due to the inverse square relationship in the Law of Universal Gravitation, the force of attraction between a planet and an object will drop to zero only at an infinite distance from the planet...

learning objectives • Express gravitational potential energy for two masses Gravitational energy is the potential energy associated with gravitational force, as work is required to elevate objects against Earth’s gravity. The potential energy due to elevated positions is called gravitational potential energy, and is evidenced by water in an elevated reservoir or kept behind a dam. If an object falls from one point to another point inside a gravitational field, the force of gravity will do positive work on the object, and the gravitational potential energy will decrease by the same amount. Consider a book placed on top of a table. As the book is raised from the floor to the table, some external force works against the gravitational force. If the book falls back to the floor, the “falling” energy the book receives is provided by the gravitational force. Thus, if the book falls off the table, this potential energy goes to accelerate the mass of the book and is converted into kinetic energy. When the book hits the floor, this kinetic energy is converted into heat and sound by the impact. The factors that affect an object’s gravitational potential energy are its height relative to some reference point, its mass, and the strength of the gravitational field it is in. Thus, a book lying on a table has less gravitational potential energy than the same book on top of a taller cupboard, and less gravitational potential energy than a heavier book lying on the same table. An object at ...

https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FBook%253A_University_Physics_(OpenStax)%2FBook%253A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)%2F13%253A_Gravitation%2F13.04%253A_Gravitational_Potential_Energy_and_Total_Energy \( \newcommand\) • • • • • • • • • • • • • Gravitational Potential Energy beyond Earth We defined \[ \Delta U = mg(y_2− y_1) \label\] Note two important items with this definition. First, \(U → 0\) as \(r → \infty\). The potential energy is zero when the two masses are infinitely far apart. Only the difference in \(U\) is important, so the choice of \(U = 0\) for \(r = \infty\) is merely one of convenience. (Recall that in earlier gravity problems, you were free to take \(U = 0\) at the top or bottom of a building, or anywhere.) Second, note that \(U\) becomes increasingly more negative as the masses get closer. That is consistent with what you learned about potential energy in Example \(\PageIndex to find the change in potential energy of the payload. That amount of work or energy must be supplied to lift the payload. Solution Paying attention to the fact that we start at Earth’s surface and end at 400 km above the surface, the change in \(U\) is We insert the values • \(m = 9000\; kg\) • \(M_\; J\; per\; month \ldotp \nonumber\] So our result is an energy expenditure equivalent to 10 months. However, this is just the energy needed to raise the...

U = − G M m R , is the Close to the Earth's surface, the gravitational field is approximately constant, and the gravitational potential energy of an object reduces to U = m g h , is given by F = G M m r 2 (for example the radius of Earth) of the two mass points, the force is integrated with respect to displacement: Gravitational Potential Energy U = − G M m R above the surface is Δ U = G M m R − G M m R + h = G M m R ( 1 − 1 1 + h / R ) . is constant, then this expression can be simplified using the Δ U ≈ G M m R [ 1 − ( 1 − h R ) ] Δ U ≈ G M m h R 2 Δ U ≈ m ( G M R 2 ) h . , this reduces to • ^ a b hyperphysics.phy-astr.gsu.edu . Retrieved 10 January 2017. • For a demonstration of the negativity of gravitational energy, see The Inflationary Universe: The Quest for a New Theory of Cosmic Origins (Random House, 1997), • MacDougal, Douglas W. (2012). Newton's Gravity: An Introductory Guide to the Mechanics of the Universe (illustrateded.). Springer Science & Business Media. p.10. 978-1-4614-5444-1. • Tsokos, K. A. (2010). Physics for the IB Diploma Full Colour (reviseded.). 978-0-521-13821-5. • Fitzpatrick, Richard (2006-02-02). farside.ph.utexas.edu. The University of Texas at Austin. • The Classical Theory of Fields, (1951), Pergamon Press,