## Gravitational Field And Gravitational Potential And Gravitational Potential Energy

The space surrounding a body where gravitational attraction can be felt due to the mass of the body, i.e., the space in which the body exerts a gravitational force on any other body, is called its gravitational field.

Actually, the gravitational field due to any object is extended up to infinity. However, the space beyond a certain distance, where the gravitational force becomes negligible in reality, may not be considered as the gravitational field of the body.

**Gravitational Field:** Along with the identification of the space that acts as the gravitational field of the body, the measurement of the field at every point of that space is also important. The physical quantity that defines the field at any point is called the gravitational field intensity, which in many cases, is mentioned only as gravitational field.

**Gravitational Field Intensity:** The gravitational force acting on a body of unit mass, placed at any point in a gravitational field, is called the gravitational field intensity of that point.

**Read and Learn More: Class 11 Physics Notes**

**Calculation Of Gravitational Field Due To A Point Mass:** Let a particle of mass M is placed at point O and a second particle of mass m is placed at point P which is at a distance r apart. The mass M creates a field \(\vec{E}\) at the position of mass m which results in a force on the mass m. The force due to the field \(\vec{E}\) is,

⇒ \(\vec{E}\) = m\(\vec{E}\)

But the force \(\vec{F}\) on the mass m due to the mass M is, \(F=\frac{G M m}{r^2} \text {, acting along } \overrightarrow{P O}\)

Thus, the gravitational field at P is, \(E=\frac{G M}{r^2} \text { along } \overrightarrow{P O}\)

In vector notation, it can be written as, \(\vec{E}=-\frac{G M}{r^2} \hat{r}\)

where \(\hat{r}\) is the unit vector along \(\overrightarrow{O P} \text {. }\).

The negative sign characterises the attractive nature of the gravitational field. It is to be noted that

- Both force and acceleration are vector quantities, and so the gravitational field intensity is also a vector quantity.
- A gravitational force can be felt at any point even due to an assembly of point masses instead of a single point mass. In that case, the gravitational field intensities due to all the masses are to be added by the vector addition method to get the total intensity,
- If the gravitational field intensity at a point, i.e., the force on a unit mass placed at that point is E, the force that will act on a mass m, placed at that point, is F = mE.

**Gravitational Field Units:**

**CGS System:**cm · s^{-2}**SI:**m · s^{-2}

**Gravitational Potential:** The work done by an external agent to bring a unit mass from infinity to a point in the gravitational field is called the gravitational potential at that point.

**Gravitational Potential Units:**

**CGS System:** erg · g^{-1}

**SI**: J · kg^{-1}

At an infinite distance, where no gravitational field exists, the gravitational potential is taken to be zero. To bring a unit mass from that point to a point in the gravitational field, work is done by the gravitational force alone.

As displacement occurs in the same direction as that of force, by convention, work done is positive. We know that change in potential or potential energy is, therefore, negative. So it can be said that, as the gravitational force is always attractive, the gravitational potential at any point in a gravitational field is negative.

**Expression For The Gravitational Potential:** Let the point O be the centre of a body of mass M, taken as the origin. The gravitational potential at a point P, in the gravitational field of the mass M, is to be determined,

Let r = distance OP, \(\hat{r}\) = unit vector along OP,

so that, \(\overrightarrow{O P}=\vec{r}=r \hat{r}\)

Now, \(\vec{E}=-\frac{G M}{r^2} \hat{r}\) = gravitational field intensity,

and \(d \vec{r}=-d r \hat{r}\) = an infinitesimal displacement from infinity towards point P.

So, the work done to displace a unit mass by \(d \vec{r}\) is,

∴ \(\vec{E} \cdot d \vec{r}=-\frac{G M}{r^2} \hat{r} \cdot(-d r \hat{r})=\frac{G M}{r^2} d r\) (because \(\hat{r} \cdot \hat{r}=1\))

The gravitational potential (V) at the point P = work done to bring a unit mass from infinity to the point P, at a distance r from M.

So, V = \(\int_{\infty}^r \vec{E} \cdot d \vec{r}=G M \int_{\infty}^r \frac{d r}{r^2}=G M\left[-\frac{1}{r}\right]_{\infty}^r\)

i.e., V=\(-\frac{G M}{r}\)

The gravitational field intensity and the potential are related as \(\vec{E}=-\frac{d V}{d r} \hat{r}\).

It is to be noted that

- As work is a scalar quantity, gravitational potential is also a scalar quantity.
- If V is the gravitational potential at a point, then the potential energy of a body of mass m kept at that point = mV.

**Gravitational Potential Energy Definition:** The gravitational potential energy of a body at a point is defined as the amount of work done in bringing the body from infinity to that point in the gravitational field.

**Gravitational Potential Energy Unit:**

**CGS System:**erg**SI:**joule

Whenever two material bodies are kept at a finite distance apart, the system possesses potential energy. It is because some work must have been done to bring them at finite distance apart from infinity.

Potential energy is positive, if the force between the two bodies is repulsive and negative if the force between them is attractive in nature. In the case of two material bodies, this potential energy is called gravitational potential energy.

**Expression For The Gravitational Potential Energy:** Let us consider that the earth is a uniform sphere of radius R and mass M. We have to calculate the gravitational potential energy of the body of mass m at a point A, such that OA = r

By definition, the gravitational potential energy of the body at point A,

U = work done in bringing the body from infinity to point A. Suppose, at any instant of time, the body is situated at the point P, such that OP = x.

Then the gravitational force on the body at the point P is given by, F = \(\frac{G M m}{x^2}\)

Now, let the body be displaced by an infinitesimally small distance dx. The small amount of work done is given by \(\)

Therefore, the total work done in bringing the body from infinity to point A,

W = \(\int_{\infty}^r \frac{G M m}{x^2} d x=G M m \int_{\infty}^r x^{-2} d x=-G M m\left[\frac{1}{x}\right]_{\infty}^r\)

= \(-G M m\left(\frac{1}{r}-\frac{1}{\infty}\right)=-\frac{G M m}{r}\)

This work is stored in the body as gravitational potential energy.

∴ \(W=U=\frac{-G M m}{r}\)

It should be noted that, as the distance r increases, the gravitational potential energy becomes less negative, i.e., increases and it becomes zero, i.e., maximum, when r = ∞

## Newtonian Gravitation And Planetary Motion – Gravitational Numerical Examples

**Example 1. Two small but heavy spheres, each of mass M are on a horizontal plane separated by a distance r. Find the gravitational potential at the mid-point of the line joining the centres of the two spheres.**

**Solution:**

Gravitational potential at the mid-point of the line and joining the two centres, \(V=\frac{-G M}{r / 2}+\frac{-G M}{r / 2}=-\frac{2 G M}{r}-\frac{2 G M}{r}=-\frac{4 G M}{r}.\)

**Example 2. At the vertices of an equilateral triangle of side a, three particles each of mass m are kept. Find the gravitational potential and the gravitational field at the centroid of the triangle.**

**Solution:**

Here AB = BC = CA = a; perpendiculars drawn from the points A, B and C on respective opposite sides meet at the centroid O and proceed to bisect the sides BC, CA and AB respectively.

Let AO = BO = CO = r

From ΔABD, AD = a sin60° = \(\frac{\sqrt{3}}{2} a\)

Also, AO = \(\frac{2}{3}\) AD

∴ \(r=\frac{2}{3} \times \frac{\sqrt{3}}{2} a=\frac{a}{\sqrt{3}} .\)

Hence the gravitational potential at the centroid O is \(V=-\frac{G m}{r}+\frac{-G m}{r}+\frac{-G m}{r}=-\frac{3 G m}{\frac{a}{\sqrt{3}}}=-\frac{3 \sqrt{3} G m}{a}\)

Let the gravitational field intensities at O due to masses at A, B and C be F_{A} (directed from O to A), F_{B} (directed from O to B) and F_{C} (directed from O to C) respectively.

∴ \(F_A=\frac{G m}{r^2}=F_B=F_C\)

Resolving the intensities along and perpendicular to the direction of AO, we get,

⇒ \(F_A-F_B \cos 60^{\circ}-F_C \cos 60^{\circ}=F_A-\frac{F_B}{2}-\frac{F_C}{2}=0\) (because \(F_B=F_C=F_A\))

and \(0+F_C \sin 60^{\circ}-F_B \sin 60^{\circ}=\frac{F_{C^{\sqrt{3}}}}{2}-\frac{F_B \sqrt{3}}{2}=0\) (because \(F_B=F_C\))

Hence, the gravitational field intensity at the centroid of the triangle = 0.

**Example 3. Find the potential energy of a system of four particles of equal masses placed at the comers of a square of side l. Also, obtain the potential at the centre of the square.**

**Solution:**

Four particles, each of mass m, are placed at the four corners of a square ABCD of side l as shown.

Here AO = CO = BO = DO = \(\frac{l}{\sqrt{2}}\)

We know that the gravitational potential energy associated with two particles of masses m_{1} and m_{2} at a distance r is -G\(\frac{m_1 m_2}{r}\)

Here, masses at the four corners are at a distance l from one another and the two diagonal pairs are at a distance \(\frac{l}{\sqrt{2}}+\frac{l}{\sqrt{2}}=\sqrt{2} l\)

So, the gravitational potential energy of the whole system,

⇒ \(E_p=4\left(-\frac{G m^2}{l}\right)+2\left(-G \frac{m^2}{\sqrt{2} l}\right)=-2 G \frac{m^2}{l}\left(2+\frac{1}{\sqrt{2}}\right)\)

= \(-5.41 \frac{G m^2}{l}\)

Gravitational potential at the centre (O) of the square is \(V=4\left(-\frac{G m}{\frac{l}{\sqrt{2}}}\right)=-4 \sqrt{2} \frac{G m}{l}\)

**Example 4. Four particles of masses m, 2m, 3m and 4m are placed at the four corners of a square of side a. Find the gravitational force on a particle of mass m placed at the centre of the square.**

**Solution:**

Let O be the centre of the square ABCD of side a.

∴ BD = \(\sqrt{a^2+a^2}=\sqrt{2} a\)

Here, OA = OB = OC = OD = \(\frac{B D}{2}=\frac{a}{\sqrt{2}}=x \text { (say). }\)

Suppose, the particle of mass m placed at O experiences gravitational forces F_{1}, F_{2}, F_{3}, and F_{4} due to the particles placed at A, B, C, and D respect

Now, \(F_1=G \frac{m \times m}{x^2}\) along \(\overrightarrow{O A}\)

⇒ \(F_2=G \frac{m \times 2 m}{x^2} \text { along } \overrightarrow{O B}\)

⇒ \(F_3=G \frac{m \times 3 m}{x^2} \text { along } \overrightarrow{O C}\)

⇒ \(F_4=G \frac{m \times 4 m}{x^2} \text { along } \overrightarrow{O D}\)

∴ The resultant of \(F_1\) and \(F_3\),

⇒ \(F^{\prime}=G \frac{m \times 3 m}{x^2}-G \frac{m \times m}{x^2}=G \frac{2 m^2}{x^2} \text { along } \overrightarrow{O C}\)

and the resultant of \(\mathrm{F}_2\) and \(\mathrm{F}_4\), \(F^{\prime \prime}=G \frac{m \times 4 m}{x^2}-G \frac{m \times 2 m}{x^2}=G \frac{2 m^2}{x^2} \text { along } \overrightarrow{O D}\)

Here, \(F^{\prime}\) and \(F^{\prime \prime}\) are equal in magnitude and at right angles to each other. The resultant of \(F^{\prime}\) and \(F^{\prime \prime}\),

F= \(\sqrt{F^{\prime 2}+F^{\prime \prime 2}}\) = \(F^{\prime}\sqrt{2}\)

= \(\frac{G .2 m^2}{x^2} \times \sqrt{2}=\frac{G .2 m^2}{\frac{a^2}{2}} \times \sqrt{2}=4 \sqrt{2} \frac{G m^2}{a^2}\)

**Example 5. What will be the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth?**

**Solution:**

Let M be the mass of the earth. Then, the gravitational potential energy of the object of mass m at the surface of the earth,

⇒ \(U_i=-\frac{G M m}{R}\)

and the potential energy of the object, when taken to a height equal to R,

⇒ \(U_f=-\frac{G M m}{R+R}=-\frac{G M m}{2 R}\)

∴ Potential energy gain = \(U_f-U_i\)

= \(-\frac{G M m}{2 R}-\left(-\frac{G M m}{R}\right)=\frac{G M m}{2 R}\)