Question

The force of attraction between two unit point masses separate by a unit distance is called
A. Gravitational potential
B. Acceleration due to gravity
C. Gravitational field
D. Universal gravitational constant

Answer 1

Hint: in this question we first need to write the definition of each option then by using the equation
$F = mg$ and $g = \dfrac{{GM}}{{{r^2}}}$ we get $F = mg = \dfrac{{mGM}}{{{r^2}}}$. Now we put M=1 and m=1 and distance r=1 and get $F = \dfrac{{G \times (1) \times (1)}}{{{1^2}}} = G$ which is the required answer.

Complete Step-by-Step solution:
We will first look at the definition of the given option one by one.

A. The gravitational potential at a position is the work done per unit mass that would be required to move a body to that position from a reference position. The reference position is that position where the potential is 0, it is conventionally at an infinitely far away from any mass.

B. The acceleration due to gravity is denoted by the symbol g and is defined as the acceleration of an object because of the force acting on that object by the gravitational field of Earth. This force is written as
$F = mg$------------------------- (1)
Here m is the mass of the object. If $m = 1$, we get
$F = g$
From this also say that g is the force on the unit mass. The g is independent of the mass of the body and can be accurately calculated in various ways.

C . The gravitational force per unit mass that is the force exerted on a mass at that point is known as the gravitational field. The gravitational field is a vector quantity and it always points in the direction of the force that the mass would feel. For an object of mass \[M\] , the measure of the gravitational field strength $g$, at distance $r$ from mass $M$ is given by
 $g = \dfrac{{GM}}{{{r^2}}}$---------------------------- (2)

D. The universal gravitational constant is a proportionality constant that is used in Newton’s Law of Universal Gravitation and it is usually denoted by G. This is different from as g denotes the acceleration due to gravity. The value of the universal gravitational constant is approximately taken as $G = 6.673 \times {10^{ - 11}}N{m^2}K{g^{ - 2}}$.
Now by using equation (1) and equation (2) we get the equation for the force due to gravitational field that is
$F = mg = \dfrac{{mGM}}{{{r^2}}}$---------------------- (3)
And put M=1 and m=1and distance r=1 in equation as shown in figure 1
We get $F = \dfrac{{G \times (1) \times (1)}}{{{1^2}}} = G$
Therefore, force of attraction between two unit point masses separate by a unit distance is the universal constant. Hence option D is correct.

Note: For solving these types of questions we need to have a clear understanding about the terms that are related to the gravitational force that is Gravitational potential, Acceleration due to gravity, Gravitational field, and Universal gravitational constant. Then we also need to be well versed with the expression related to each term.