Outside the Solid Sphere

The force field that exists in space around every mass or group of masses is known as a gravitational field. This field extends in all ways, but the magnitude of the gravitational force reduces as one gets farther away from the object. It is measured in newtons per kilogramme (N/kg) units of force per mass.

 A gravitational field is a form of force field that is analogous to electric and magnetic fields for electrically charged particles and magnets. In this article, we will read about the gravitational field outside the spherical shell through the gravitational field outside the spherical shell. 

Newton’s law of Gravitation

According to the law, every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. The force is directed along the line that connects the particle lines.

 Consider two masses, m1 and m2, with their centres separated by r. Newton’s law of gravitation infers the gravitational force between them.

The vector form

F =𝐺𝑚1𝑚2/r²  𝑟̂

F12 denotes force on mass 1 due to mass 2, and F21 denotes force on mass 2 due to mass 1.

 It’s worth noting that F12 and F21 are both equal and opposite. An action-reaction pair is formed by gravitational force between two particles.

Finding Gravitational field In Solid Sphere

Case I 

The gravitational field outside the solid sphere. Let M be the sphere’s mass and R be its radius. We must compute the gravitational field at P.

∫ 𝑑𝐼 = ∫𝐺𝑑𝑚/𝑟²

∫ 𝑑𝐼 =𝐺/r² = ∫ 𝑑𝑚 =𝐺𝑚/𝑟²

As a result, a uniform sphere can be treated as a single particle of equal mass placed at its centre to calculate the gravitational field at an external point.

Case 2 

The gravitational field at the internal point of a uniform sphere. Assume point P is inside the solid sphere; in this case, rR is the sphere that can be divided into thin spherical shells, all of which are centred at O. If the mass of such a shell is dm, then the gravitational field about this spherical shell will be- 

𝑑𝐼 =𝐺𝑑𝑚/𝑟²

𝑎𝑙𝑜𝑛𝑔 𝑃𝑂

 ∫ 𝑑𝐼 = ∫𝐺𝑑𝑚/𝑟²

∫ 𝑑𝐼 =𝐺/r² ∫ 𝑑𝑚

But dm = density × volume

∫ 𝑑𝑚 =𝑀4/3 𝜋𝑅³/4/3 πr³

         = 𝑀r³/ 𝑅³

∴ 𝐼 = 𝐺𝑀/𝑅³

As a result, the gravitational field produced by a uniform sphere at an internal point is proportional to the distance of the point from the sphere’s centre. The field is zero at r = 0’s centre.

r = R at the sphere’s surface

I = GM/R²

The gravitational field due to a solid sphere is continuous but not easily measurable. The gravitational field at a point within the sphere is only due to the mass enclosed by the surface passing through the point, which is shown by the shaded portion in the diagram, and the field due to the external volume is zero.

𝐼 = ∫ 𝑑𝐼=𝐺r² ∫ 𝑑𝑚 =𝐺𝑀/ r²

Electric Fuel Due To Spherical Shell

In this case, when a point is inside a spherical shell, the surface crossing through it does not enclose any mass, so I = 0.

Point P is located outside of the spherical shell.

𝐼 = ∫ 𝑑𝐼 =𝐺/𝑟²

 ∫ 𝑑𝑚 =𝐺𝑀/𝑟²

The gravitational field produced by a thin spherical shell is a discontinuous and non-differentiable component.

Solved Examples Of Gravitational Field Outside The Spherical Shell

Question

A satellite weighing 1000 kilogrammes is in synchronous orbit around the Earth. The period of this synchronous orbit corresponds to the rotation of the earth around its axis, which is assumed to be 24 hours, giving the satellite the appearance of being stationary. 

a) What is the satellite’s orbital radius?

b) What is the satellite’s altitude?

c) What is the satellite’s kinetic energy?

Solution

1. Let M represent the mass of the planet and m represent the mass of the satellite. When a satellite orbits, the universal gravitational force, and the centripetal force are equal.

G M m / R2 = m v2 / R, where v is the satellite’s orbital speed and R is its orbital radius.

v = 2πR / T

G M m / R2= m (2πR / T)2 / R

 Solve to obtain: R3 = M G T2 / (4π2)

 The altitude h of the satellite is given by the radius of the Earth, which is 6371 km.

h = 42,211 – 6371 = 35,840 km

3. The satellite’s kinetic energy Ek is given by

Ek = (1/2) m v² = (1/2) 1000 (2πR / T)² = (1/2) 1000 (2π × 42,211,000 / (24 × 60 × 60))² = 4.7 ×109 J

Conclusion

 A gravitational field is present in any region where an atom is subject to a force solely determined by the atom’s mass and position. This field is a vector quantity. In this article, we have learned about the gravitational field outside the spherical shell. We have Solved Examples Of Gravitational Field Outside The Spherical Shell to understand the topic clearly.