Physics Formulas

Gravitational Field Formula

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, Ā 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 š‘”, is given by: š¹ = mš‘”. At the surface of the Earth, š‘” has a magnitude of \(\frac{GM}{R^{2}_{E}}= 9.81 ms^{-2}\), where \(R_{E}\) is the radius of the Earth.

Gravitational Field Formula

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 ()

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

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 Ɨ 1024 kg.

Application of theĀ 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 forĀ 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 \(9.799 \frac{m}{s^{2}}\)

Share with friends

Customize your course in 30 seconds

Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
tutor
tutor
Ashhar Firdausi
IIT Roorkee
Biology
tutor
tutor
Dr. Nazma Shaik
VTU
Chemistry
tutor
tutor
Gaurav Tiwari
APJAKTU
Physics
Get Started

5
Leave a Reply

avatar
5 Comment threads
0 Thread replies
7 Followers
 
Most reacted comment
Hottest comment thread
5 Comment authors
Yashdeep tiwariKwame DavidumerMalek safrinRoger Carmichael Recent comment authors
  Subscribe  
newest oldest most voted
Notify of
Roger Carmichael
Guest

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

Malek safrin
Guest

I have realy intrested to to this topic

umer
Guest
umer

m=f/a correct this

B. Akshaya
Guest
B. Akshaya

M=f/g

Kwame David
Guest

Interesting studies

Yashdeep tiwari
Guest
Yashdeep tiwari

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

Customize your course in 30 seconds

Which class are you in?
No thanks.