# 4.4 Gravitational field  (Page 3/3)

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## Example

Problem 1: A charged particle of mass “m” carries a charge “q”. It is projected upward from Earth’s surface in an electric field “E”, which is directed downward. Determine the nature of potential energy of the particle at a given height, “h”.

Solution : The charged particle is acted upon simultaneously by both gravitational and electrostatic fields. Here, gravity works against displacement. The work by gravity is, therefore, negative. Hence, potential energy arising from gravitational field (with reference from surface) is positive as :

${U}_{G}=-{W}_{G}=-\left(-{F}_{G}h\right)=m{E}_{G}h=mgh$

As given in the question, the electrostatic field is acting downward. Since charge on the particle is positive, electrostatic force acts downward. It means that work by electrostatic force is also negative. Hence, potential energy arising from electrostatic field (with reference from surface) is :

${U}_{E}=-{W}_{E}=-\left(-{F}_{E}h\right)=q{E}_{E}h=qEh$

Total potential energy of the charged particle at a height “h” is :

$⇒U={U}_{G}+{U}_{E}=\left(mgh+qEh\right)=\left(mg+qE\right)h$

The quantities in the bracket are constant. Clearly, potential energy is a function of height.

It is important to realize that description in terms of respective fields enables us to calculate forces without referring to either Newton's gravitation law or Coulomb's law of electrostatic force.

## Gravitational field due to a point mass

Determination of gravitational force strength due to a point mass is easy. It is so because, Newton's law of gravitation provides the expression for determining force between two particles.

Let us consider a particle of mass, "M", for which we are required to find gravitational field strength at a certain point, "P". For convenience, let us consider that the particle is situated at the origin of the reference system. Let the point, where gravitational field is to be determined, lies at a distance "r" from the origin on the reference line.

We should make it a point to understand that the concept of gravitational field is essentially "one" particle/ body/entity concept. We need to measure gravitational force at the point, "P", on a unit mass as required by the definition of field strength. It does not exist there. In order to determine field strength, however, we need to visualize as if unit mass is actually present there.

We can do this two ways. Either we visualize a point mass exactly of unit value or we visualize any mass, "m", and then calculate gravitational force. In the later case, we divide the gravitational force as obtained from Newton's law of gravitation by the mass to get the force per unit mass. In either case, we call this point mass as test mass. If we choose to use a unit mass, then :

$⇒E=F=\frac{GMX1}{{r}^{2}}=\frac{GM}{{r}^{2}}$

On the other hand, if we choose any arbitrary test mass, "m", then :

$⇒E=\frac{F}{m}=\frac{GMm}{{r}^{2}m}=\frac{GM}{{r}^{2}}$

However, there is a small catch here. The test mass has its own gravitational field. This may unduly affect determination of gravitational field due to given particle. In order to completely negate this possibility, we may consider a mathematical expression as given here, which is more exact for defining gravitational field :

$E=\underset{m\to 0}{\overset{}{\mathrm{lim}}}\phantom{\rule{1em}{0ex}}\frac{F}{m}$

Nevertheless, we know that gravitational force is not a very strong force. The field of a particle of unit mass can safely be considered negligible.

The expression for the gravitational field at point "P", as obtained above, is a scalar value. This expression, therefore, measures the magnitude of gravitational field - not its direction. We can realize from the figure shown above that gravitational field is actually directed towards origin, where the first particle is situated. This direction is opposite to the positive reference direction. Hence, gravitational field strength in vector form is preceded by a negative sign :

$⇒E=\frac{F}{m}=-\frac{GM}{{r}^{2}}\stackrel{‸}{r}$

where " $\stackrel{‸}{r}$ " is unit vector in the reference direction.

The equation obtained here for the gravitational field due to a particle of mass, "M", is the basic equation for determining gravitational field for any system of particles or rigid body. The general idea is to consider the system being composed of small elements, each of which can be treated at particle. We, then, need to find the net or resultant field, following superposition principle. We shall use this technique to determine gravitational field due to certain regularly shaped geometric bodies in the next module.

## Example

Problem 2 : The gravitational field in a region is in xy-plane is given by 3 i + j . A particle moves along a straight line in this field such that work done by gravitation is zero. Find the slope of straight line.

Solution : The given gravitational field is a constant field. Hence, gravitational force on the particle is also constant. Work done by a constant force is given as :

$W=F.r$

Let "m" be the mass of the particle. Then, work is given in terms of gravitational field as :

$⇒W=mE.r$

Work done in the gravitational field is zero, if gravitational field and displacement are perpendicular to each other. If “ ${s}_{1}$ ” and “ ${s}_{2}$ ” be the slopes of the direction of gravitational field and that of straight path, then the slopes of two quantities are related for being perpendicular as :

${s}_{1}{s}_{2}=-1$

Note that slope of a straight line is usually denoted by letter “m”. However, we have used letter “s” in this example to distinguish it from mass, which is also represented by letter “m”.

In order to find the slope of displacement, we need to know the slope of the straight line, which is perpendicular to the direction of gravitational field.

Now, the slope of the line of action of gravitational field is :

$⇒{s}_{1}=\frac{1}{3}$

Hence, for gravitational field and displacement to be perpendicular,

$⇒{s}_{1}{s}_{2}=\left(\frac{1}{3}\right){s}_{2}=-1$

${s}_{2}=-3$

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