In physics science a gravitational field is the influence that a massive body extends into the space around itself, producing a force on another massive body. So in other terms, a gravitational field helps us to explain the gravitational phenomena and is measured in newtons per kilogram (N/kg). Sir Isaac Newton discovered gravity and its forces. The gravitational field varies slightly at the earth’s surface. Let us study gravitational field formula in detail.

**What is Gravitational Field?**

The gravitational force per unit mass that would be exerted on a small test mass at that point is defined as the gravitational field. It is a vector field, and points in the direction of the force that a small test mass would feel at that point. Let’s consider a point particle of mass M, then the magnitude of the resultant gravitational field strength denoted by term g, at a distance of r, from M, is given by the formula,

\(g= \frac{GM}{r^{2}}\)

Newton’s Law of Gravitation states that the gravitational force between two point masses M and m a distance r apart in a vacuum, is attractive, acts along the line joining their centres, and is proportional to the masses and inversely proportional to the square of their separations. This is the formula,

\(F\alpha \frac{Mm}{r^{2}}\)

In the SI system, the constant of proportionality is G, the gravitational constant, which has a value of \(6.67 \times 10^{-11}Nm^{2} kg^{-2}\), and so we may write this as

\(F= \frac{GMm}{r^{2}}\)

The gravitational force or the weight acting on a mass m, in the gravitational field 𝑔, is given by: 𝐹 = m𝑔. At the surface of the Earth, 𝑔 has a magnitude of \(\frac{GM}{R^{2}_{E}}= 9.81 ms^{-2}\), where \(R_{E}\) is the radius of the Earth.

Source: en.wikipedia.org

**The Formula of Gravitational Field:**

Near the earth the acceleration due to gravity depends on the distance of an object from earth’s center. The gravitational field formula is very useful. Using it helps to find the field strength, meaning the acceleration due to gravity at any position around the Earth. The radius of the Earth is \(R_{E}= 6.38 \times 10^{6}m\) , and so values of r in the formula are (typically) greater than this radius. The strength of gravitational field is measured in Newtons per kilogram.

\(\left ( \frac{N}{kg} \right )\), or in the same units as acceleration is \(\frac{m}{s^{2}}\)

\(g\left ( r \right )=\frac{Gm_{g}}{r^{2}}\)

g(r) = Earth’s gravitational field strength, \(\left ( \frac{N}{kg} \right ) or \frac{m}{s^{2}}\)

G = gravitational constant ()

\( m_{E}\)= mass of the Earth \(\left ( 5.98\times 10^{24}\right ) kg\)

r = distance from the center of the Earth (m).

The gravitational pull between two objects only affects their motion when at least one of the objects is very massive. Earth has a mass of about 6 × 1024 kg.

**Application of the Gravitational Field Formula**

Gravitational field calculation has a lot of uses in the aerospace segment. It is in use for positioning satellites in space. Also used while sending rockets in space.

**Solved Examples for Gravitational Field Formula**

Q.1. What is the gravitational field strength at the surface of the Earth, \(R_{g}= 6.38 \times 10^{6}m\)?

Ans- The gravitational field strength at the surface of the Earth is:

\(g\left ( r \right )= \frac{Gm_{g}}{r^{2}}\)

\(g\left ( R_{E} \right ) = \frac{Gm_{E}}{\left ( R_{E} \right )^{2}}\)

= \(\frac{\left (6.67 \times 10^{-11} N.\frac{m^{2}}{kg^{2}} \right )\left ( 5.98\times 10^{24}kg \right )}{ \left ( 6.38 \times 10^{6}m \right )^{2}}\)

\(= \frac{\left (6.67 \times 10^{-11} N.\frac{m^{2}}{kg^{2}} \right )\left ( 5.98\times 10^{24}kg \right )}{ 40.7044\times 10^{12}m^{2}}\)

= \(0.9799 \times 10^{-11+24-12}\frac{N}{kg}\)

\(= 0.9799 \times 10^{1}\frac{N}{kg}\)

\(g\left ( R_{E} \right ) = 0.9799 \frac{N}{kg}\)

The gravitational field strength at the Earth’s surface is approximately \(9.799 \frac{N}{kg}\). This is equivalent to an acceleration due to gravity at the Earth’s surface of \(9.799 \frac{m}{s^{2}}\)

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

Interesting studies

It is already correct f= ma by second newton formula…