Universal Law of Gravitation Questions in English

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force.
The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
For two objects of masses $m_1$ and $m_2$ and the distance between them $r$,the force $(F)$ of attraction acting between them is given by the universal law of gravitation as:
$F = \frac{G m_1 m_2}{r^2}$
Where,$G$ is the universal gravitation constant and its value is $6.67 \times 10^{-11} \text{ Nm}^2 \text{ kg}^{-2}$.

(N/A) Let $M_E$ be the mass of the Earth and $m$ be the mass of an object on its surface. If $R$ is the radius of the Earth,then according to the universal law of gravitation,the gravitational force $(F)$ acting between the Earth and the object is given by the relation:
$F = \frac{G M_E m}{R^2}$
Where:
$G$ is the universal gravitational constant,
$M_E$ is the mass of the Earth,
$m$ is the mass of the object,
$R$ is the distance between the center of the Earth and the object (which is equal to the radius of the Earth).

(D) According to the universal law of gravitation,the gravitational force $(F)$ acting between two objects is inversely proportional to the square of the distance $(r)$ between them,i.e.,
$F = G \frac{m_{1} m_{2}}{r^{2}}$
If the distance $r$ is reduced to $r/2$,the new gravitational force $(F')$ will be:
$F' = G \frac{m_{1} m_{2}}{(r/2)^{2}}$
$F' = G \frac{m_{1} m_{2}}{r^{2}/4}$
$F' = 4 \times (G \frac{m_{1} m_{2}}{r^{2}})$
$F' = 4F$
Therefore,if the distance is reduced to half,the gravitational force becomes four times the original value.

(C) According to Newton's universal law of gravitation,every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This force is mutual. Therefore,the Earth attracts the moon with a force that is exactly the same as the force with which the moon attracts the Earth. This is in accordance with Newton's third law of motion,which states that for every action,there is an equal and opposite reaction.

(N/A) According to the universal law of gravitation,the force of gravitation between two objects is given by $F = \frac{G M m}{r^{2}}$.
$(i)$ $F$ is directly proportional to the masses of the objects. If the mass of one object is doubled,then the gravitational force will also get doubled.
$(ii)$ $F$ is inversely proportional to the square of the distance between the objects. If the distance is doubled,the gravitational force becomes one-fourth of its original value. Similarly,if the distance is tripled,the gravitational force becomes one-ninth of its original value.
$(iii)$ $F$ is directly proportional to the product of the masses of the objects. If the masses of both objects are doubled,the new force $F' = \frac{G(2M)(2m)}{r^{2}} = 4 \times \frac{G M m}{r^{2}} = 4F$. Thus,the gravitational force becomes four times the original value.

(C) According to the universal law of gravitation,the force of attraction between the Earth and the Sun is given by:
$F = \frac{G \times M_{\text{Sun}} \times M_{\text{Earth}}}{R^2}$
Where:
$M_{\text{Sun}} = 2 \times 10^{30} \, kg$
$M_{\text{Earth}} = 6 \times 10^{24} \, kg$
$R = 1.5 \times 10^{11} \, m$
$G = 6.7 \times 10^{-11} \, Nm^2 \, kg^{-2}$
Substituting the values:
$F = \frac{6.7 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{(1.5 \times 10^{11})^2}$
$F = \frac{80.4 \times 10^{43}}{2.25 \times 10^{22}}$
$F = 35.73 \times 10^{21} \, N = 3.57 \times 10^{22} \, N$
Thus,the gravitational force is $3.57 \times 10^{22} \, N$.

(D) Let the two stones meet after a time $t$.
For the stone dropped from the tower:
Initial velocity,$u = 0\, m/s$.
Let the displacement of the stone in time $t$ from the top of the tower be $s$.
Acceleration due to gravity,$g = 9.8\, m/s^2$.
Using the equation of motion,$s = ut + \frac{1}{2}at^2$:
$s = 0 \times t + \frac{1}{2} \times 9.8 \times t^2$
$s = 4.9t^2$ ............. $(1)$
For the stone thrown upwards:
Initial velocity,$u = 25\, m/s$.
Let the displacement of the stone from the ground in time $t$ be $s'$.
Acceleration due to gravity,$g = -9.8\, m/s^2$.
Using the equation of motion,$s' = ut + \frac{1}{2}at^2$:
$s' = 25t - \frac{1}{2} \times 9.8 \times t^2$
$s' = 25t - 4.9t^2$ ............. $(2)$
The sum of the displacements of both stones at the meeting point is equal to the total height of the tower,$100\, m$:
$s + s' = 100$
$4.9t^2 + 25t - 4.9t^2 = 100$
$25t = 100$
$t = 4\, s$
In $4\, s$,the distance covered by the falling stone from the top is:
$s = 4.9 \times (4)^2 = 4.9 \times 16 = 78.4\, m$.
Therefore,the stones will meet after $4\, s$ at a height of $(100 - 78.4) = 20.6\, m$ from the ground.

(C) According to Newton's law of universal gravitation,the force $F$ between two objects of masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
When the masses of both objects are halved,the new masses become $m_1' = \frac{m_1}{2}$ and $m_2' = \frac{m_2}{2}$.
The new gravitational force $F'$ is $F' = G \frac{(\frac{m_1}{2})(\frac{m_2}{2})}{r^2} = G \frac{m_1 m_2}{4r^2} = \frac{1}{4} F$.
Therefore,the gravitational force becomes $F/4$.

(B) The relation $F = G \frac{Mm}{d^2}$ represents Newton's Law of Universal Gravitation.
In this formula,$G$ is known as the Universal Gravitational Constant.
It is called 'universal' because its value remains constant throughout the universe,regardless of the location,the nature of the masses,or the medium between them.
Its value is approximately $6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ everywhere.
Therefore,option $B$ is the correct statement.

(C) Newton's law of universal gravitation states that 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 the distance between their centers.
Mathematically,it is expressed as $F = G \frac{m_1 m_2}{r^2}$,where $F$ is the gravitational force,$G$ is the universal gravitational constant,$m_1$ and $m_2$ are the masses of the two bodies,and $r$ is the distance between them.
Since this law applies to all objects with mass,it describes the gravitational force between any two bodies having some mass in the universe.

(D) The quantity $G$ represents the Universal Gravitational Constant.
By definition,a universal constant has the same value throughout the universe,regardless of the location,the mass of the objects,or the distance between them.
Therefore,the value of $G$ is independent of the mass and radius of the earth.

(A) According to Newton's law of universal gravitation,the force $F$ between two objects of masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$.
If the mass of each particle is doubled,the new masses become $m_1' = 2m_1$ and $m_2' = 2m_2$.
The new gravitational force $F'$ is given by $F' = G \frac{(2m_1)(2m_2)}{r^2} = 4 \times (G \frac{m_1 m_2}{r^2}) = 4F$.
Therefore,the gravitational force becomes $4$ times the original value.

(B) The atmosphere is a layer of gases surrounding the Earth.
These gas molecules have mass and are subject to the Earth's gravitational pull.
Gravity acts as the force that keeps these gases close to the Earth's surface,preventing them from escaping into outer space.
Therefore,the atmosphere is held to the Earth by gravity.

(C) According to Newton's law of universal gravitation,the force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = G \frac{m_1 m_2}{r^2}$,where $G$ is the universal gravitational constant.
If $m_1 = 1 \text{ unit}$,$m_2 = 1 \text{ unit}$,and $r = 1 \text{ unit}$,then $F = G \frac{1 \times 1}{1^2} = G$.
Therefore,the force of attraction between two unit point masses separated by a unit distance is numerically equal to the universal gravitational constant $G$.

(N/A) The source of the centripetal force required for a planet to revolve around the Sun is the gravitational force exerted by the Sun on the planet.
According to Newton's law of universal gravitation,this force $F$ is given by the formula $F = G \frac{M \times m}{r^2}$,where $M$ is the mass of the Sun,$m$ is the mass of the planet,$r$ is the distance between them,and $G$ is the universal gravitational constant.
Therefore,the force depends on the product of the masses of the planet and the Sun $(M \times m)$ and is inversely proportional to the square of the distance $(r^2)$ between them.

(C) The gravitational force of the Sun provides the necessary centripetal force to keep the Earth in its orbit.
This force acts perpendicular to the velocity of the Earth,constantly changing its direction to maintain a circular path.
Because the gravitational pull is exactly balanced by the requirements of the circular motion,the Earth neither falls into the Sun nor escapes into space.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses $(M_1 \times M_2)$ and inversely proportional to the square of the distance $(r^2)$ between their centres.
Mathematically, this is expressed as $F = G \frac{M_1 M_2}{r^2}$, where $F$ is the gravitational force, $G$ is the universal gravitational constant, $M_1$ and $M_2$ are the masses of the two objects, and $r$ is the distance between them.

(A) Yes,gravitational forces obey Newton's third law of motion.
According to Newton's third law,for every action,there is an equal and opposite reaction.
In the case of gravitation,if object $A$ exerts a gravitational force on object $B$,then object $B$ simultaneously exerts an equal and opposite gravitational force on object $A$.

(A) The gravitational force between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by Newton's Law of Gravitation: $F = G \frac{m_1 m_2}{r^2}$.
This force depends only on the masses of the objects and the distance between them.
The gravitational constant $G$ is a universal constant and does not depend on the medium between the masses.
Therefore,the gravitational force remains the same regardless of whether the objects are in air or any other medium.
Thus,the ratio of the force in air to the force in any other medium is $1: 1$.

(C) No,it is not possible to shield a body from a gravitational field.
This is because the gravitational force is a fundamental interaction that acts between all masses in the universe.
Unlike electric or magnetic fields,which can be blocked by certain materials (like Faraday cages for electric fields),the gravitational force is independent of the nature of the medium present between two bodies.
Therefore,no material can act as a shield against gravity.

(N/A) One common phenomenon related to the universal law of gravitation is the motion of planets around the Sun. According to Newton's law of universal gravitation,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 the distance between them. This gravitational force provides the necessary centripetal force that keeps planets in their respective orbits around the Sun.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses ($M_1$ and $M_2$) and inversely proportional to the square of the distance $(d^2)$ between their centers.
Mathematically,this is expressed as: $F = G \frac{M_1 M_2}{d^2}$,where $F$ is the gravitational force and $G$ is the universal gravitational constant.

(C) If the force of gravity disappears suddenly,all objects will lose the centripetal force required to maintain their circular motion around the Earth. According to Newton's first law of motion,objects will continue to move in a straight line tangent to their current path at the point where gravity vanished.

(N/A) $(i)$ According to Newton's law of universal gravitation,every object in the universe attracts every other object with a force that is proportional to the product of their masses.
$(ii)$ The earth exerts a gravitational force on the stone,causing it to accelerate towards the earth.
$(iii)$ The stone also exerts an equal and opposite gravitational force on the earth.
$(iv)$ However,due to the extremely large mass of the earth compared to the stone,the acceleration produced in the earth is negligible,making its motion towards the stone imperceptible.

(N/A) $(i)$ The stone falls towards the earth because of the gravitational force exerted by the earth on the stone. According to the universal law of gravitation,every object in the universe attracts every other object with a force. Since the earth is massive,its gravitational pull on the stone is significant,causing the stone to accelerate towards it.
$(ii)$ According to the universal law of gravitation,the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Since the masses of ordinary objects in a room are very small,the gravitational force between them is extremely weak. This force is negligible compared to the frictional forces and the gravitational pull of the earth,which is why they do not move towards each other.

(N/A) $1$. The gravitational force exerted by the Sun on the planets provides the necessary centripetal force for their motion along a circular orbit.
$2$. The direction of this force is always towards the center of the orbit (i.e.,towards the Sun).
$3$. If this force were to disappear suddenly,the planet would no longer be able to maintain its circular path and would continue to move in a straight line along the tangent to its orbit at that point,as per Newton's first law of motion.

(B) The formula for the force of gravitation is $F = G \frac{m_1 m_2}{r^2}$.
Given:
$m_1 = 10 \; kg$
$m_2 = 20 \; kg$
$r = 10 \; m$
$G = 6.67 \times 10^{-11} \; Nm^2 kg^{-2}$
Substituting the values into the formula:
$F = \frac{6.67 \times 10^{-11} \times 10 \times 20}{(10)^2} \; N$
$F = \frac{6.67 \times 10^{-11} \times 200}{100} \; N$
$F = 6.67 \times 10^{-11} \times 2 \; N$
$F = 13.34 \times 10^{-11} \; N$

(C) According to Newton's law of universal gravitation,the force of attraction between two objects is given by $F = G \frac{M_1 M_2}{r^2}$.
This force is mutual,meaning the earth exerts a force on the lead,and the lead exerts an equal and opposite force on the earth.
According to Newton's third law of motion,for every action,there is an equal and opposite reaction.
Therefore,the gravitational force exerted by the earth on $1 \,kg$ of lead is exactly equal in magnitude to the gravitational force exerted by the $1 \,kg$ of lead on the earth.

(N/A) According to Newton's law of universal gravitation,every object in the universe attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between them $(F = G \frac{M_1 M_2}{r^2})$.
While all bodies do exert a gravitational force on each other,the magnitude of this force is extremely small for objects of ordinary size because their masses are negligible compared to the mass of the Earth.
In contrast,the Earth has an enormous mass. Therefore,the gravitational force exerted by the Earth on any object near its surface is significantly larger than the mutual gravitational forces between those objects.
Consequently,the Earth's gravitational pull dominates,causing all objects to be attracted towards the centre of the Earth.

(N/A) $(i)$ The centripetal force required for the earth to revolve around the sun is provided by the gravitational force of attraction between the sun and the earth.
$(ii)$ According to the law of universal gravitation,the magnitude of this force depends on the mass of the sun $(M)$,the mass of the earth $(m)$,and the square of the distance $(r^2)$ between their centers. The formula is $F = G \frac{Mm}{r^2}$,where $G$ is the universal gravitational constant.

(N/A) Let $m_{1}$ and $m_{2}$ be the masses of the two bodies and $M$ be the mass of the earth.
Let the distance of the two bodies from the centre of the earth be $r$.
According to Newton's law of universal gravitation,the force $F$ exerted by the earth on a body of mass $m$ at a distance $r$ is given by $F = \frac{GMm}{r^{2}}$,where $G$ is the universal gravitational constant.
For body $A$,the force is $F_{1} = \frac{GMm_{1}}{r^{2}}$.
For body $B$,the force is $F_{2} = \frac{GMm_{2}}{r^{2}}$.
Given that the forces are equal,$F_{1} = F_{2}$.
Therefore,$\frac{GMm_{1}}{r^{2}} = \frac{GMm_{2}}{r^{2}}$.
By cancelling the common terms $\frac{GM}{r^{2}}$ from both sides,we get $m_{1} = m_{2}$.
Thus,the masses of the two bodies are equal.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically,$F = G \frac{M \times m}{d^2}$,where $F$ is the gravitational force,$M$ and $m$ are the masses of the two objects,and $d$ is the distance between them.
The $SI$ unit of the universal gravitational constant $G$ is $N \ m^2 \ kg^{-2}$.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically,$F = G \frac{m_1 m_2}{r^2}$,where $G$ is the universal gravitational constant.
If the distance $r$ is tripled,the new distance becomes $r' = 3r$.
The new force $F'$ will be $F' = G \frac{m_1 m_2}{(3r)^2} = G \frac{m_1 m_2}{9r^2} = \frac{1}{9} F$.
Therefore,the force becomes $\frac{1}{9}$ times the original force.

(N/A) $(i)$ It is the weakest force in nature.
$(ii)$ It is always attractive in nature and acts between any two objects having mass.

(N/A) $(i)$ Newton's law of gravitation is applicable to all objects in the universe,regardless of their size,mass,or distance,hence it is called the universal law of gravitation.
$(ii)$ The universal law of gravitation is significant because it successfully explains the following phenomena:
$(a)$ The gravitational force that binds us to the Earth.
$(b)$ The motion of the Moon around the Earth.
$(c)$ The motion of planets around the Sun.
$(d)$ The occurrence of tides in the oceans due to the gravitational pull of the Moon and the Sun.

(N/A) $(i)$ The force of attraction between two objects is directly proportional to the product of their masses, i.e., $F \propto m_{1} m_{2}$.
The force of attraction between two objects is inversely proportional to the square of the distance between them, i.e., $F \propto \frac{1}{r^{2}}$.
$(ii)$ Given:
Mass of the Earth, $M = 6 \times 10^{24} \, kg$
Mass of the satellite, $m = 6.5 \times 10^{20} \, kg$
Distance between them, $r = 3.35 \times 10^{6} \, km = 3.35 \times 10^{9} \, m$
Gravitational constant, $G = 6.7 \times 10^{-11} \, Nm^{2} kg^{-2}$
Using the formula $F = G \frac{Mm}{r^{2}}$:
$F = \frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 6.5 \times 10^{20}}{(3.35 \times 10^{9})^{2}}$
$F = \frac{261.3 \times 10^{33}}{11.2225 \times 10^{18}}$
$F \approx 23.28 \times 10^{15} \, N$ or $2.33 \times 10^{16} \, N$.

(N/A) The gravitational force between two objects is given by the expression $F = \frac{G m_1 m_2}{r^2}$.
$(a)$ If the mass of one object is doubled $(m_1' = 2m_1)$,the new force $F' = \frac{G(2m_1)m_2}{r^2} = 2F$. Thus,the force doubles.
$(b)$ If the distance between the objects is doubled $(r' = 2r)$,the new force $F' = \frac{G m_1 m_2}{(2r)^2} = \frac{G m_1 m_2}{4r^2} = \frac{1}{4}F$. Thus,the force becomes one-fourth of the original force.
$(c)$ If the masses of both objects are doubled ($m_1' = 2m_1$ and $m_2' = 2m_2$),the new force $F' = \frac{G(2m_1)(2m_2)}{r^2} = 4 \times \frac{G m_1 m_2}{r^2} = 4F$. Thus,the force becomes four times the original force.

(N/A) The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The $SI$ unit of the universal gravitational constant $G$ is $N \, m^2 \, kg^{-2}$.
Given:
Initial force $F_1 = 100 \, N$
Final force $F_2 = 50 \, N$
We know that the gravitational force $F$ is given by $F = \frac{G m_1 m_2}{r^2}$,which implies $F \propto \frac{1}{r^2}$.
Therefore,$\frac{F_1}{F_2} = \left(\frac{r_2}{r_1}\right)^2$.
Substituting the values:
$\frac{100}{50} = \left(\frac{r_2}{r_1}\right)^2$
$2 = \left(\frac{r_2}{r_1}\right)^2$
$\frac{r_2}{r_1} = \sqrt{2}$
Thus,the distance between the objects should be increased by a factor of $\sqrt{2}$ (approximately $1.414$ times the original distance) to make the force $50 \, N$.

(N/A) Newton's law of gravitation is called a universal law because it is applicable to all objects in the universe,regardless of their mass,size,shape,or the distance between them. It describes the gravitational force acting between any two particles of matter anywhere in the universe.

(B) The gravitational force between two small objects on the surface of the Earth is extremely weak because their masses are very small.
In contrast,the mass of the Earth is enormous.
Due to this,the gravitational attraction between any object and the Earth is significantly stronger than the mutual attraction between the objects themselves.
Consequently,all objects are pulled towards the center of the Earth rather than towards each other.

(C) No,it is not possible to shield a body from gravitational effects.
This is because gravitational interaction is a fundamental force that acts between all objects with mass.
Unlike electromagnetic forces,which can be shielded by placing a material between the charges,gravitational force does not depend on the nature of the intervening medium.
Therefore,gravity cannot be blocked or neutralized by any known material.

(C) The gravitational force exerted by the earth on the moon provides the necessary centripetal force required for the moon to move in a circular orbit around the earth.
This gravitational force acts perpendicular to the direction of the moon's velocity,which changes the direction of the moon's motion continuously without changing its speed.
As a result,the moon keeps revolving in its orbit instead of falling towards the earth.

(A) The universal law of gravitation is given by the formula $F = \frac{G m_1 m_2}{r^2}$.
Given:
Mass of the first object,$m_1 = 50 \, kg$.
Mass of the second object,$m_2 = 120 \, kg$.
Distance between the objects,$r = 10 \, m$.
Gravitational constant,$G = 6.673 \times 10^{-11} \, N \, m^2 \, kg^{-2}$.
Substituting these values into the formula:
$F = \frac{6.673 \times 10^{-11} \times 50 \times 120}{10^2}$
$F = \frac{6.673 \times 10^{-11} \times 6000}{100}$
$F = 6.673 \times 10^{-11} \times 60$
$F = 400.38 \times 10^{-11} \, N$
$F = 4.0038 \times 10^{-9} \, N \approx 4.0 \times 10^{-9} \, N$.

(D) Given:
Mass of Jupiter,$M_1 = 1.9 \times 10^{27} \, kg$
Mass of Sun,$M_2 = 1.99 \times 10^{30} \, kg$
Distance between Jupiter and Sun,$r = 7.8 \times 10^{11} \, m$
Gravitational constant,$G = 6.67 \times 10^{-11} \, N \cdot m^2 \cdot kg^{-2}$
Using the universal law of gravitation formula,$F = \frac{G M_1 M_2}{r^2}$:
$F = \frac{6.67 \times 10^{-11} \times 1.9 \times 10^{27} \times 1.99 \times 10^{30}}{(7.8 \times 10^{11})^2}$
$F = \frac{6.67 \times 1.9 \times 1.99 \times 10^{46}}{60.84 \times 10^{22}}$
$F \approx 4.146 \times 10^{23} \, N$
Rounding to the nearest provided option,the gravitational force is $4.4 \times 10^{23} \, N$.

(A) According to Newton's law of universal gravitation,the force $F$ is inversely proportional to the square of the distance $r$ between the objects: $F \propto \frac{1}{r^2}$.
Given initial force $F_1 = 100 \, N$ and final force $F_2 = 50 \, N$.
We have the ratio: $\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}$.
Substituting the values: $\frac{50}{100} = \frac{r_1^2}{r_2^2} \implies \frac{1}{2} = \frac{r_1^2}{r_2^2}$.
Taking the square root on both sides: $\frac{1}{\sqrt{2}} = \frac{r_1}{r_2}$.
Therefore,$r_2 = \sqrt{2} \, r_1$.
Thus,the distance must be increased by a factor of $\sqrt{2}$.

(C) According to Newton's law of universal gravitation,the gravitational force $F$ between two bodies of masses $m_1$ and $m_2$ separated by a distance $r$ is given by the formula:
$F = G \frac{m_1 m_2}{r^2}$
Here,$G$ is the universal gravitational constant.
From this formula,it is clear that the force $F$ is inversely proportional to the square of the distance $r$ between the two bodies.
Therefore,$F \propto \frac{1}{r^2}$.

(A) According to Newton's first law of motion,an object in motion will continue to move in a straight line at a constant speed unless acted upon by an external force.
Since the earth moves in a nearly circular orbit around the sun,its direction of motion is constantly changing at every point.
$A$ change in direction implies a change in velocity,which constitutes acceleration.
According to Newton's second law,acceleration requires a net force.
Therefore,the revolution of the earth around the sun serves as evidence that a centripetal force (gravitational force) must be acting on the earth,directed towards the sun.

(D) The gravitational constant $(G)$ is a universal constant.
It is defined as the force of attraction between two unit masses separated by a unit distance.
Its value is $6.674 \times 10^{-11} \ N \ m^2/kg^2$.
Since it is a universal constant,it does not depend on the masses of the objects,the distance between them,the temperature of the medium,or any other physical conditions.
Therefore,the correct option is $D$.

(A) The universal law of gravitation,expressed as $F = G \frac{m_1 m_2}{r^2}$,describes the gravitational force between two point masses.
For extended objects,this formula is strictly valid only when the objects are spherical and their mass distribution is uniform (spherically symmetric).
In such cases,the entire mass of the sphere can be considered to be concentrated at its center,allowing the distance $r$ to be measured between the centers of the two spheres.

(D) According to Newton's law of universal gravitation,the force $F$ between two objects of masses $m_1$ and $m_2$ separated by a distance $r$ is given by the formula: $F = G \frac{m_1 m_2}{r^2}$.
Here,$G$ is the universal gravitational constant.
If the distance $r$ is doubled,the new distance becomes $r' = 2r$.
The new gravitational force $F'$ will be: $F' = G \frac{m_1 m_2}{(2r)^2} = G \frac{m_1 m_2}{4r^2}$.
This can be rewritten as: $F' = \frac{1}{4} \times (G \frac{m_1 m_2}{r^2}) = \frac{1}{4} F$.
Therefore,the gravitational force becomes one-fourth of its original value.