Gravitational Potential Energy | Derivation, Formula, Numerical, FAQ & MCQs

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Definition of Gravitational Potential Energy

Gravitational potential energy of a body at a point is defined as the work done in bringing the body from infinity to that point in the gravitational field of another body (such as Earth). It is denoted by U.

Derivation of Gravitational Potential Energy

Gravitational Potential Energy

Consider a body of mass m placed at a distance x from the centre of Earth of mass M.

According to Newton's law of gravitation, the gravitational force acting on the body is:

$$ F=\frac{GMm}{x^2} $$

Here,

  • G = Universal Gravitational Constant
  • M = Mass of Earth
  • m = Mass of the body
  • x = Distance of the body from Earth's centre

Let the body be displaced through a very small distance dx.

The small work done during this displacement is:

$$ dW=F\,dx $$

Substituting the value of gravitational force:

$$ dW=\frac{GMm}{x^2}dx $$

To find the total work done in bringing the body from infinity to distance r, integrate both sides:

$$ W=\int_{\infty}^{r}\frac{GMm}{x^2}dx $$

Since G, M and m are constants, they can be taken outside the integral:

$$ W=GMm\int_{\infty}^{r}\frac{dx}{x^2} $$

Writing \(\frac{1}{x^2}\) as \(x^{-2}\):

$$ W=GMm\int_{\infty}^{r}x^{-2}dx $$

Using the integration formula:

$$ \int x^n dx=\frac{x^{n+1}}{n+1} $$

For \(n=-2\):

$$ \int x^{-2}dx = \frac{x^{-1}}{-1} = -\frac{1}{x} $$

Substituting this result:

$$ W=GMm\left[-\frac{1}{x}\right]_{\infty}^{r} $$

Applying the limits:

$$ W = GMm \left[ -\frac{1}{r} - \left( -\frac{1}{\infty} \right) \right] $$

Since

$$ \frac{1}{\infty}=0 $$

Therefore,

$$ W = GMm \left( -\frac{1}{r} +0 \right) $$

$$ W=-\frac{GMm}{r} $$

Since gravitational potential energy is equal to the work done in bringing the body from infinity to that point,

$$ U=W $$

Hence,

$$ \boxed{U=-\frac{GMm}{r}} $$

Discussion

  1. Gravitational potential energy increases with increase in distance r from the centre of Earth, i.e., it becomes less negative.
  2. At infinity, gravitational potential energy is zero. $$ U=0 \quad \text{at} \quad r=\infty $$
  3. Gravitational potential energy is always negative because gravitational force is attractive.

Change in Gravitational Potential Energy

Let a body of mass m move from a distance r₁ to another distance r₂ from the centre of Earth of mass M.

The gravitational potential energy at distance r₁ is:

$$ U_1=-\frac{GMm}{r_1} $$

The gravitational potential energy at distance r₂ is:

$$ U_2=-\frac{GMm}{r_2} $$

Change in gravitational potential energy is given by:

$$ \Delta U=U_2-U_1 $$

Substituting the values of \(U_2\) and \(U_1\):

$$ \Delta U= \left(-\frac{GMm}{r_2}\right) - \left(-\frac{GMm}{r_1}\right) $$

Removing the brackets:

$$ \Delta U= -\frac{GMm}{r_2} +\frac{GMm}{r_1} $$

Taking the common factor \(GMm\):

$$ \Delta U= GMm \left( \frac{1}{r_1} - \frac{1}{r_2} \right) $$

Multiplying by \((-1)\) inside the bracket:

$$ \Delta U= -GMm \left( \frac{1}{r_2} - \frac{1}{r_1} \right) $$

Hence,

$$ \boxed{ \Delta U= -GMm \left( \frac{1}{r_2} - \frac{1}{r_1} \right) } $$

When a body moves from distance \(r_1\) to \(r_2\),

$$ \Delta U = -GMm \left( \frac{1}{r_2} - \frac{1}{r_1} \right) $$

Since $r_1$ > $r_2$ , so change in Gravitation P.E of the body is negative. It means when a body is brought near to the earth , it's P.E decrease.

Potential Energy Near Earth's Surface

Let a body of mass m be raised from the Earth's surface to a height h.

Initial distance from the centre of Earth:

$$ r_1=R $$

Final distance from the centre of Earth:

$$ r_2=R+h $$

Using the formula for change in gravitational potential energy:

$$ \Delta U = -GMm \left( \frac{1}{r_2} - \frac{1}{r_1} \right) $$

Substituting \(r_1=R\) and \(r_2=R+h\):

$$ \Delta U = -GMm \left( \frac{1}{R+h} - \frac{1}{R} \right) $$

Taking LCM of the two fractions:

$$ \Delta U = -GMm \left( \frac{R-(R+h)} {R(R+h)} \right) $$

Simplifying the numerator:

$$ R-(R+h) = R-R-h = -h $$

Therefore,

$$ \Delta U = -GMm \left( \frac{-h} {R(R+h)} \right) $$

Negative sign cancels with negative sign:

$$ \Delta U = \frac{GMmh} {R(R+h)} $$

Dividing numerator and denominator by \(R\):

$$ \Delta U = \frac{GMm}{R} \left( \frac{h/R} {1+h/R} \right) $$

For heights close to Earth's surface,

$$ h \ll R $$

Using the binomial approximation:

$$ \frac{1}{1+\frac{h}{R}} \approx 1-\frac{h}{R} $$

Therefore,

$$ \Delta U = \frac{GMm}{R} \left[ 1- \frac{1}{1+\frac{h}{R}} \right] $$

$$ \Delta U = \frac{GMm}{R} \left[ 1- \left( 1-\frac{h}{R} \right) \right] $$

$$ \Delta U = \frac{GMm}{R} \left( \frac{h}{R} \right) $$

$$ \Delta U = \frac{GMmh}{R^2} $$

Since,

$$ g=\frac{GM}{R^2} $$

Substituting \(g\) in the above equation:

$$ \Delta U=mgh $$

Hence,

$$ \boxed{\Delta U=mgh} $$


Important Facts

  • Gravitational force varies inversely as the square of distance. $$ F\propto \frac{1}{r^2} $$
  • Gravitational potential energy varies inversely as distance. $$ U\propto -\frac{1}{r} $$
  • Potential energy on Earth's surface: $$ U=-\frac{GMm}{R} $$
  • As distance increases, potential energy becomes less negative.
  • At infinity, gravitational potential energy becomes zero.

Graph of Gravitational Potential Energy

Graph of Gravitational Potential Energy

  • At Earth's surface: $$ U=-\frac{GMm}{R} $$
  • As \(r\) increases, the value of \(U\) increases.
  • At infinity: $$ U\rightarrow 0 $$
  • The graph approaches zero asymptotically.

Potential Energy of a System of Particles

For a system of three particles:

$$ U = -G \left( \frac{m_1m_2}{r_{12}} + \frac{m_2m_3}{r_{23}} + \frac{m_3m_1}{r_{31}} \right) $$


Frequently Asked Questions (FAQ)

Q1. What is gravitational potential energy?

Gravitational potential energy is the work done in bringing a body from infinity to a point in a gravitational field.

Q2. Why is gravitational potential energy negative?

Gravitational force is attractive. Therefore, gravitational potential energy is negative.

Q3. What is the SI unit of gravitational potential energy?

The SI unit of gravitational potential energy is Joule (J).

Q4. What is the dimensional formula of gravitational potential energy?

$$ [ML^2T^{-2}] $$

Q5. What is the gravitational potential energy at infinity?

$$ U=0 $$

Q6. What is the gravitational potential energy on Earth's surface?

$$ U=-\frac{GMm}{R} $$

Q7. When is \(\Delta U=mgh\) valid?

It is valid when the height \(h\) is very small compared to Earth's radius (\(h \ll R\)).

Q8. How does gravitational potential energy vary with distance?

$$ U\propto -\frac{1}{r} $$


Quiz (MCQs)

  1. The gravitational potential energy of a body at infinity is:

    A) Positive   B) Negative   C) Zero   D) Infinite

    Answer: C) Zero

  2. The expression for gravitational potential energy at a distance \(r\) is:

    A) GMm/r   B) -GMm/r   C) mgh   D) GMm/r²

    Answer: B) -GMm/r

  3. The SI unit of gravitational potential energy is:

    A) Newton   B) Watt   C) Joule   D) Pascal

    Answer: C) Joule

  4. Gravitational potential energy is generally:

    A) Positive   B) Zero   C) Negative   D) Infinite

    Answer: C) Negative

  5. Near Earth's surface, change in gravitational potential energy is:

    A) mg   B) mgh   C) GMm/R   D) GMm/R²

    Answer: B) mgh

  6. Gravitational potential energy varies as:

    A) r   B) r²   C) 1/r²   D) 1/r

    Answer: D) 1/r

  7. The dimensional formula of potential energy is:

    A) [MLT⁻²]   B) [ML²T⁻²]   C) [M⁰L²T⁻²]   D) [ML⁻¹T⁻²]

    Answer: B) [ML²T⁻²]

  8. If distance from Earth increases, gravitational potential energy:

    A) Decreases   B) Remains constant   C) Becomes more negative   D) Increases

    Answer: D) Increases

  9. The relation \(g=\frac{GM}{R^2}\) is used to derive:

    A) Kinetic Energy   B) Work-Energy Theorem   C) mgh   D) Escape Velocity

    Answer: C) mgh

  10. The gravitational potential energy of a system equals:

    A) Sum of kinetic energies   B) Sum of pairwise potential energies   C) Product of masses   D) Zero

    Answer: B) Sum of pairwise potential energies

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