Gravitational Field Intensity Due to a Ring

A gravitational field can be defined as the influence that a huge body has on the space surrounding it. It can also be called the force that one body exerts on another heavy body. The gravitational field is calculated in newtons per kilogram (N/kg). It also helps in explaining gravitational phenomena. Gravity and its forces were discovered by Sir Isaac Newton. At the Earth’s surface, the gravitational field changes moderately. In this article, we will discuss gravitational field intensity due to a ring. 

Gravitational field due to a ring

Let us assume you want to determine the gravitational field at a point “P” that exists on the central axis of the ring of mass “M” and radius “a”. We can assume a tiny mass “dm” lies on the circular ring. The gravitational field is because of this elemental mass and is along with PA. Thus, its magnitude can be expressed as:

The gravitational field can be seen on axial point “P”.

dM = Gdm / PA² 

= Gdm / (a²+ b²) 

You can solve this gravitational field which is in the parallel direction and perpendicular to the axis in the plane of OAP.

The net gravitational field here is axial.

dEI = dE cos ∅ 

Here, you can observe three things. First, you can observe from the figure that measurements of “y” and “θ” are the same for all masses. Also, you can assume that they have equal elemental masses. Thus, you can say that the magnitude of the gravitational field because of y elements of mass “dm” will be the same. It occurs because they are at an equal distance from point “P”.

Secondly, the pairs of elemental masses are on symmetrically opposite sides of the ring. The perpendicular sections of elemental field intensity will be directed in opposing directions. These perpendicular elements will now add up to zero for the complete ring when integration is done. If the mass distribution on the ring is uniform, you can say that zero-field strength is perpendicular to the axial line. The net gravitational intensity for a uniform ring can be calculated because integration is done on the axial components of elemental field strength. 

So the mathematical expression can be represented as follows: 

The elements which are perpendicular will cancel each other out.

E = ∫ dE cos∅

E = ∫ Gdm cos ∅ / (a ² + b²) 

The trigonometric ratio “cosθ” will be constant at any point on the ring. You can take out the cosine ratio and other constants from the integral,

E =G cos∅/ (a² + b²) ∫ dm 

Now integrating, m= 0 to m = M, 

E =G M cos∅/ (a² + b²)

For triangle OAP, 

Cos∅ = r / (a² + b²)½

Substituting for cos ∅ in the given equation,

E = GMr / (a² + b² ) 3/2 

E = 0 when r = 0. Hence, it can be inferred that the centre of the ring contains no gravitational force. This is also expected since the gravitational forces produced by two opposed identical elemental masses are equal, which are opposing and balancing each other.

Maximum gravitational field position

By differentiating its expression concerning linear distance and equating it to zero, you can determine the maximum value of the gravitational field.

dE /dr = 0

This gives us 

r = a /√ 2 

Substituting it in the given expression of the gravitational field, the maximum field strength due to a circular ring is :

Emax = GMr /[2½ (a²+ a²/2)3/2 ]

= GMr / 3 √ 3a²

Question: Determine the gravitational field strength at the surface of the Earth, which is given as Rg=6.38×106m?

Answer: The gravitational field strength at the surface of the Earth can be determined by using the formula: 

g ( r) = Gmg / r²

g ( RE ) = Gm g / ( RE )²

( 6.67 x 10-11 N. m² / kg²) ( 5.98 x 10²⁴ kg) / ( 6.38 x 10⁶ m)² 

( 6.67 x 10-11 N. m²/ kg²) (5.98 x10²⁴ kg) / 40. 7044 x 10¹² m² 

= 0.9799 x 10-11+24-12 N/ kg 

= 0.9799 x 10¹ N/kg

= g (RE ) = 0.9799 N/kg 

Therefore, from the above question, we can determine gravitational field strength at the Earth’s surface, which is approximately 9.799. It is also equivalent to an acceleration that happens due to the gravity of the Earth’s surface at 9.799 m/s².

Why do we use the gravitational field formula? 

Gravitational field calculation has many applications in aerospace engineering. The formula can be used for positioning satellites in space. It can also be Also used for sending space shuttles into space. 

Conclusion

Therefore, we can determine the gravitational field intensity due to the gravitational force per unit mass that would be exerted on a small test mass at that point is defined as the gravitational field. The gravitational field is a vector quantity and points in the direction of the force that a small test mass would feel at that point. In this article, we learned about gravitational field intensity due to ring and gravitational field intensity due to ring importance.