a 10 kg mass is lifted to a height of 2 m what is its potentail energy at this position

Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it tin can be said that gravitational potential energy is an energy that is related to gravitational strength or to gravity.

The most common example that can assist you understand the concept of gravitational potential energy is if you take ii pencils. One is placed at the table and the other is held above the tabular array. Now, we can country that the pencil which is high volition have greater gravitational potential energy than the pencil that is at the table.

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What we can learn hither is that the pencil or whatever object, in particular, volition take the potential to do work because of its location in the gravitational field. We will learn about the topic in detail below.

Table of Content:

• Gravitational Potential Energy
• Formula
• Derivation
• Gravitational Potential
• Intensity vs Potential
• G.P of Point Mass
• Spherical Shell
• Solid Sphere
• Solved Examples

What is Gravitational Potential Free energy?

When a body of mass (k) is moved from infinity to a point inside the gravitational influence of a source mass (M) without accelerating it, the amount of work done in displacing it into the source field is stored in the course of potential energy. This is known as gravitational potential free energy. Information technology is represented with the symbol Ug.

Explanation: We know that the potential energy of a body at a given position is defined as the energy stored in the body at that position. If the position of the body changes due to the application of external forces the change in potential energy is equal to the corporeality of work done on the body past the forces.

Under the action of gravitational force, the piece of work washed is independent of the path taken for a change in position so the forcefulness is a conservative force. Also, all such forces have some potential in them.

The gravitational influence on a trunk at infinity is nada, therefore, potential free energy is zero, which is called a reference bespeak.

Gravitational Potential Energy – Video Lesson

Gravitational Potential Free energy Formula

The equation for gravitational potential free energy is:

⇒ GPE = m⋅grand⋅h

Where,

• g is the mass in kilograms
• yard is the dispatch due to gravity (nine.8 on Earth)
• h is the height above the ground in meters

⇒ Also Read:

• Gravitational Forcefulness
• Kepler's Laws
• Gravitational Field Intensity

Derivation of Gravitational Potential Energy Equation

Consider a source mass 'M' is placed at a point forth the x-axis, initially, a test mass 'grand' is at infinity. A pocket-size amount of work done in bringing it without dispatch through a very small distance (dx) is given by

dw = Fdx

Hither, F is an attractive force and the displacement is towards the negative x-axis management and so F and dx are in the same direction. And so,

dw = (GMm/x^two)dx

Integrating on both sides

\(\brainstorm{assortment}{l}due west = \int_{\infty }^{r} \frac{GMm}{x^{ii}}dx\finish{array} \)

\(\begin{assortment}{50}west = -[\frac{GMm}{x}]_{\infty }^{r}\end{array} \)

\(\begin{array}{l}w = -[\frac{GMm}{r}] – (\frac{-GMm}{\infty })\finish{array} \)

\(\begin{array}{l}w = \frac{-GMm}{r}\end{array} \)

Since the work done is stored every bit its potential free energy U, therefore gravitational potential free energy at a signal which is at a distance 'r' from the source mass is given by;

U = -GMm/r

If a examination mass moves from a point inside the gravitational field to the other point within the same gravitational field of source mass, so the change in potential energy of the test mass is given by;

ΔU = GMm (one/r_i – 1/r_f)

If r_i > r_f and then ΔU is negative.

⇒ Check:Dispatch due to Gravity

Expression for Gravitational Potential Energy at Meridian (h) – Derive ΔU = mgh

If a body is taken from the surface of the earth to a point at a elevation 'h' in a higher place the surface of the world, so r_i= R and r_f= R + h then,

ΔU = GMm [1/R – 1/(R+h)]

ΔU = GMmh/R(R + h)

When, h<<R, then, R + h = R and g = GM/Rⁱⁱ.

On substituting this in the above equation nosotros get,

Gravitational Potential Energy ΔU = mgh

⇒ Note:

• The weight of a body at the heart of the globe is zero due to the fact that the value of m at the eye of the globe is nothing.
• At a point in the gravitational field where the gravitational potential energy is zippo, the gravitational field is zero.

What is Gravitational Potential?

The amount of piece of work done in moving a unit of measurement examination mass from infinity into the gravitational influence of source mass is known as gravitational potential.

Simply, information technology is the gravitational potential energy possessed by a unit test mass

⇒ V = U/m

⇒ Five = -GM/r

⇒ Important Points:

• The gravitational potential at a bespeak is always negative, Five is maximum at infinity.
• The SI unit of gravitational potential is J/Kg.
• The dimensional formula is G⁰LⁱⁱT^-ii.

⇒ Dimensional Formulas of Physical Quantities

Relation between Gravitational Field Intensity and Gravitational Potential

Integral Form:

V =

\(\begin{array}{l}-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

( If E is given and V has to be constitute using this formula)

Differential Form:

E = -dV/dr (If V is given and Eastward has to be found using this formula)

\(\begin{array}{l}\overrightarrow{~East}=\frac{\fractional 5}{\partial x}\hat{i}+\frac{\fractional V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\end{array} \)

(components along x, y and z directions).

Gravitational Potential of a Point Mass

Consider a betoken mass M, the gravitational potential at a distance 'r' from it is given by;

5 = – GM/r.

Gravitational Potential of a Spherical Crush

Consider a thin uniform spherical shell of the radius (R) and mass (Thousand) situated in space. Now,

Case 1: If indicate 'P' lies Inside the spherical shell (r<R):

Equally E = 0, V is a abiding.

The value of gravitational potential is given by, V = -GM/R.

Instance 2: If point 'P' lies on the surface of the spherical shell (r=R):

On the surface of the globe, E = -GM/R².

Using the relation

\(\begin{array}{l}5=-\mathop{\int }\vec{E}.\overrightarrow{dr}\terminate{assortment} \)

over a limit of (0 to R) we get,

Gravitational Potential (V) = -GM/R.

Case 3: If betoken 'P' lies o utside the spherical trounce (r>R):

Outside the spherical shell, E = -GM/r².

Using the relation

\(\begin{array}{l}Five=-\mathop{\int }\vec{Eastward}.\overrightarrow{dr}\end{array} \)

over a limit of (0 to r) nosotros get,

V = -GM/r.

Gravitational Potential of a Uniform Solid Sphere

Consider a thin compatible solid sphere of the radius (R) and mass (G) situated in infinite. Now,

Example 1: If betoken 'P' lies Within the uniform solid sphere (r < R):

Inside the uniform solid sphere, E = -GMr/R³.

Using the relation

\(\begin{array}{l}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

over a limit of (0 to r).

The value of gravitational potential is given by,

5 = -GM [(3R² – r^two)/2R²]

Example two: If signal 'P' lies o n the surface of the uniform solid sphere ( r = R ):

On the surface of a uniform solid sphere, E = -GM/R². Using the relation

\(\brainstorm{array}{fifty}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

over a limit of (0 to R) nosotros get,

5 = -GM/R.

Case 3: If betoken 'P' lies o utside the uniform solid sphere ( r> R):

Using the relation over a limit of (0 to r) we get, V = -GM/R.

Case four: Gravitational potential at the middle of the solid sphere is given by V =(-3/2) × (GM/R).

Gravitational Self Free energy

The gravitational self-free energy of a body is divers as the piece of work washed by an external agent in assembling the body from the infinitesimal elements that are initially at an infinite distance apart.

Gravitational Self Energy of a system of 'northward' particles:

Let us consider n particle system in which particles interact with each other at an average altitude 'r' due to their mutual gravitational allure, there are n(n – i)/2 such interactions and the potential energy of the organization is equal to the sum of the potential energy of all pairs of particles i.e.,

\(\begin{array}{l}{{U}_{south}}=\frac{1}{ii}Gn\left( n-one \right)\frac{{{m}^{ii}}}{{{r}^{2}}}\stop{array} \)

Solved Problems

Example 1. Calculate the gravitational potential energy of a body of mass 10Kg and is 25m above the ground.

Solution:

Given, Mass m = ten Kg and Height h = 25 thousand

1000.P.Due east is given as,

U = one thousand × grand × h = x Kg nine.8 m/s² × 25 m = 2450 J.

Example two. If the mass of the world is v.98 ×10²⁴ Kg and the mass of the sun is 1.99 × ten^thirty Kg and the earth is 160 million Kms abroad from the sun. Calculate the GPE of the earth.

Solution:

Given, the mass of the Earth (k) = 5.98 × 10²⁴Kg and mass of the Sun (M) = 1.99 × 10³⁰ Kg

The gravitational potential energy is given by:

U = -GMm/r

U = (half dozen.673 ∗ 10^-11 ∗ 5.98 ∗ ten²⁴ ∗1.99∗ten³⁰)/(160∗ten⁹) = 4963 x 10^xxxJ

Example iii. A basketball game weighing two.2 kg falls off a building to the ground 50 m below. Calculate the gravitational potential energy of the ball when it arrives below.

Solution:

GPE = (2.2 kg)(ix.viii m/s²)(50 chiliad) = 1078 J.

Example 4: A 2 kg body gratis falls from rest from a height of 12 m. Determine the work done by the force of gravity and the change in gravitational potential energy. Consider the acceleration due to gravity to be 10 chiliad/south².

Solution:

Since, W = mgh

Substituting the values in the above equation, we get

W = 2 × 12 × 10 = 240 N

The change in gravitational potential free energy is equal to the work washed past gravity.

Therefore, Gravitational Potential Free energy= 240 Joule.

⇒ Likewise Read:

• HC Verma
• HC Verma Solutions Vol 1
• HC Verma Solutions Vol 2

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