Solved Examples on Universal Law of Gravitation

Example 1: Two bodies of masses 5 kg and 6 kg are placed with their centres 64 m apart. Calculate the initial acceleration of two masses assuming no other forces act on them.

Solution:

Given,

Mass of first body, m₁ = 5 kg

Mass of second body, m₂ = 6 kg

Distance between two bodies, r = 64 m

Universal Gravitational Constant, G = 6.67 × 10^-11 Nm² kg^-2

F = G × (m₁ × m₂) / r² 

Substitute the given values in the above expression as,

F = 6.67 × 10^-11 Nm² kg^-2 × (5 kg × 6 kg) / (64 m)² 

   = 48.85 × 10^-14 N

Now, it is known that, the net force on an object is given by,

F = ma

a = F / m

a₁ = 48.85 × 10^-14 N / 5 kg

    = 9.77 × 10^-14 m/s²

a₂ = 48.85 × 10^-14 N / 6 kg

    = 8.142 × 10^-14 m/s²

Thus, initial accelerations of two bodies are 9.77 × 10^-14 m/s² and 8.142 × 10^-14 m/s² respectively.

Example 2: A body of mass 1 kg and another body of mass 10 kg are separated by a distance of 100 m, and they are attracted by a force of 19.6×10^-10 N. Calculate the value of the universal gravitation constant for the given case.

Solution:

Mass of the first body, m₁ = 1 kg

Mass of the second body, m₂ = 10 kg

Distance between the two bodies, r = 100 m

Gravitational Force of Attraction between the two bodies F = 19.6 × 10^-10 N.

Therefore, from the Newton’s law of Gravitation,

G = F × r² / (m₁ × m₂)

   = 19.6 × 10^-10 N × (100 m)² / (1 kg × 100 kg)

   = 19.6 × 10^-7 Nm² kg^-2 

Thus, value of universal gravitation constant for this case is 19.6 × 10^-7 Nm² kg^-2.

Example 3: If the two objects attract each other with a gravitational force of F units. If the mass of both objects was tripled and the distance between the objects also doubled. What would be the new force of attraction between two objects?

Solution:

Let the force of attraction between two bodies be F N

F = G × (m₁ × m₂) / r² 

where 
m₁ is the mass of first object
m₂ is the mass of second object
r is the distance between the two bodies

Let F₁ be the new force of attraction between two objects.

When the mass of both objects are tripled and the distance between the objects is doubled then the force acting between two objects is,

F₁ = G × (3 × m₁ × 3 × m₂) / (2r)²

    = 9G × m₁ × m₂ / 4r²

    = 9F/4 

Thus, new force of attraction between two objects is 9F/4 units.

Example 4: Calculate the resultant force acting on mass m which is at position 1 and all the masses which are equal to m are placed in the vertices of the equilateral triangle having the length of its side is r.

Solution:

Given that, the masses whose weight equal to m are placed on the three corners of equilateral triangle.

Let F₁ be the force acting on mass m which is at position 1.

By the Principle of Superposition of Gravitational Forces,

 F₁ = Forces acting on mass m due to another mass m which is at position 2 + Forces acting on mass m due to another mass m which is at position 3                       

Here 
 and  makes an angle of 60° to each other since they act along the sides of equilateral triangle.

Let 

F_res=

     =

F_res= √3f = √3×Gm²/r²

Thus the resultant force acting on mass m which is at position 1 and all the masses which are equal to m are placed in the vertices of the equilateral triangle having the length of its side is √3Gm²/r².

Example 5: Three uniform spheres each having a mass M and radius a are kept in such a way that each touches the other two. Find the magnitude of the gravitational force on any of the spheres due to the other two spheres.

Solution:

Let A, B and C be the centers of three uniform spheres.

Let  be the Force acting on C due to A and   be the force acting on C due to B

Let the total force acting on C due to A and B be 

By the Principle of Superposition of Gravitational Forces, 

  ……(1)

On joining A,B and C, we get a equilateral triangle i.e, it makes an angle of 60° to each other.

Given radius of each sphere is a

so, AB = BC = CA = 2a

From Newton’s law of Gravitation, 

Here, 

From (1),

     = 

     = 

|F| = 

Thus the magnitude of the gravitational force on any of the sphere due to the other two spheres is 

Example 6: If Mercury, Venus, and the Sun are aligned in a right-angled triangle. Calculate the vector sum of forces on venus due to both mercury and Sun. What is the direction and magnitude of the resulting force?

Solution:

Distance between Sun and Venus (r_v) = 108×10⁹ meters.

Distance between Sun and Mercury(r_m) = 57.6×10⁹ meters

Distance between Mercury and Venus (r_mv)=

r_mv = 1.08×10¹¹ meters.

Mass of Sun = 1.99×10³⁰ kg

Mass of Mercury = 3.3×10²³ kg

Mass of Venus = 4.87×10²⁴ kg

Let Force acting on Venus due to Sun is F_S

 From Newton’s law of Gravitation,

  = 6.67×10^-11×1.99×10³⁰×4.87×10²⁴/(108×10⁹)^2

F_S = 5.54×10²² N 

Let Force acting on Venus due to Mercury is F_M

 

   = (6.67 × 10^-11 × 3.3 × 10²³ × 4.87 ×10²⁴)/(1.08×10¹¹)²

F_M = 9.19×10¹⁵ N

Here Force due to Sun is more than a million times greater than Force due to Mercury and so the net Force is mainly due to Sun and its magnitude is equal to Force acting due to Sun. And the direction of resulting Force similar to direction due to Sun.

Universal Law of Gravitation or Newton’s law of Universal Gravitation as the name suggests is given by Sir Isaac Newton. This law helps us to understand the motion of very large bodies in the universe. According to this law, an attractive force always acts between two bodies that have masses. The strength of the force is directly proportional to the mass of the object and is inversely proportional to the square of the distance between them.

In this article, you are going to read about everything related to Universal Law of Gravitation including its definition, what Gravitational Law states, weight vs. Gravitational Force, etc.