Mix Examples - GRAVITATION Questions in English

(D) The universal law of gravitation is significant because:
$1$. It binds us to the Earth.
$2$. It is responsible for the motion of the Moon around the Earth.
$3$. It is responsible for the motion of planets around the Sun.
$4$. It causes the occurrence of tides due to the Moon and the Sun.

(N/A) According to Newton's law of universal gravitation,the force of attraction $(F)$ between two bodies is directly proportional to the product of their masses ($m_{1}$ and $m_{2}$) and inversely proportional to the square of the distance $(d)$ between them: $F \propto m_{1} m_{2}$ and $F \propto \frac{1}{d^{2}}$.
This hypothesis is incorrect. Two bricks tied together will fall at the same rate as a single brick under the action of gravity in a vacuum (free-fall). This is because the acceleration due to gravity $(g)$ is independent of the mass of the falling object. Since both objects experience the same acceleration $(g \approx 9.8 \ m/s^{2})$,they will reach the ground at the same time.

(N/A) Yes, the statement is true. According to Newton's $Third$ $Law$ of $Motion$, the Earth exerts a gravitational force on the apple, and the apple exerts an equal and opposite force on the Earth. According to Newton's $Second$ $Law$ of $Motion$ $(F = ma)$, the acceleration produced is inversely proportional to the mass of the object $(a = F/m)$. Since the mass of the Earth is extremely large compared to the mass of the apple, the acceleration produced in the Earth is negligible and therefore not noticeable.

(A) According to Newton's law of universal gravitation,$F = G \frac{m_1 m_2}{r^2}$.
$(i)$ If the distance $r$ is reduced to half,i.e.,$r' = r/2$,the new force $F'$ is:
$F' = G \frac{m_1 m_2}{(r/2)^2} = G \frac{m_1 m_2}{r^2 / 4} = 4 \left( G \frac{m_1 m_2}{r^2} \right) = 4F$.
So,the force becomes $4$ times the original force.
$(ii)$ If the mass of one object becomes four times,i.e.,$m_1' = 4m_1$,the new force $F''$ is:
$F'' = G \frac{(4m_1) m_2}{r^2} = 4 \left( G \frac{m_1 m_2}{r^2} \right) = 4F$.
So,the force becomes $4$ times the original force.

(N/A) $(i)$ Weightlessness is observed in a satellite because the satellite and all objects inside it are in a state of free fall towards the Earth. Since there is no external contact force (normal force) acting on the body to oppose gravity,the apparent weight becomes zero.
$(ii)$ Given:
Initial velocity $(u) = 15 \, m s^{-1}$
Final velocity at maximum height $(v) = 0 \, m s^{-1}$
Acceleration due to gravity $(g) = -9.8 \, m s^{-2}$
Using the third equation of motion: $v^2 - u^2 = 2gh$
$0^2 - (15)^2 = 2 \times (-9.8) \times h$
$-225 = -19.6 \times h$
$h = \frac{225}{19.6} \approx 11.48 \, m$
Thus,the maximum height attained is approximately $11.48 \, m$.

(N/A) This is because the gravitational force of attraction between the Sun and the Earth provides the necessary centripetal force for the Earth's orbital motion.
$(b)$ Given: Mass $m = 500 \, g = 0.5 \, kg$,Height $h = 5 \, m$,Acceleration due to gravity $g = 10 \, m s^{-2}$,Initial velocity $u = 0$.
Using the third equation of motion: $v^2 - u^2 = 2gh$
$v^2 - 0 = 2 \times 10 \times 5 = 100$
$v = \sqrt{100} = 10 \, m s^{-1}$
Momentum $p = m \times v$
$p = 0.5 \, kg \times 10 \, m s^{-1} = 5 \, kg \, m s^{-1}$.

(N/A) $(i)$ No,the friend will not agree. Weight is the force with which the earth attracts an object,given by the formula $W = m \times g$. Since the acceleration due to gravity $(g)$ is greater at the poles than at the equator,the weight of the same amount of gold will be less at the equator compared to the poles.
$(ii)$ According to Newton's third law of motion,the moon exerts an equal and opposite force on the earth. However,the earth does not move towards the moon because the mass of the earth is extremely large compared to the mass of the moon. According to Newton's second law $(F = m \times a)$,the acceleration produced is inversely proportional to the mass $(a = F / m)$. Thus,the acceleration of the earth is negligible.

(A) $G$ is the Universal Gravitational Constant with a fixed value of $6.673 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$. It remains constant everywhere in the universe.
$g$ is the acceleration due to gravity,which varies depending on the location. Its $SI$ unit is $\text{m s}^{-2}$.
$(b)$ No,the value of $g$ is not the same everywhere on Earth. The Earth is not a perfect sphere; it is flattened at the poles and bulges at the equator. Since $g = \frac{GM}{R^2}$,the radius $R$ is smaller at the poles compared to the equator. Therefore,the value of $g$ is greater at the poles and smaller at the equator.
$(c)$ According to Newton's Law of Universal Gravitation,$F = \frac{G M m}{r^2}$,so $F \propto \frac{1}{r^2}$.
If the distance $r$ is tripled $(r' = 3r)$,the new force $F'$ becomes $F' = \frac{G M m}{(3r)^2} = \frac{G M m}{9r^2} = \frac{1}{9} F$.
Thus,the gravitational force becomes one-ninth of its original value.

(B) In the experiment to measure the force required to move a wooden block,it is essential that the force applied by the spring balance is horizontal to ensure that the entire force contributes to overcoming the static friction.
In setup $A$,the string is pulled horizontally,ensuring the force vector is parallel to the surface.
In setup $B$,the string is also horizontal,but the connection is direct and stable.
In setup $C$,the use of a coiled spring can introduce vertical components of force or oscillations,leading to inaccuracies.
Setup $A$ provides the most direct and horizontal application of force,minimizing errors due to alignment,thus yielding the most accurate result.

(N/A) Let the masses of two bodies be $m_{1}$ and $m_{2}$.
Let the mass of the Earth be $M$ and the distance from the center of the Earth be $d$.
According to Newton's law of universal gravitation,the force $F$ is given by $F = \frac{GMm}{d^{2}}$.
For the two bodies,$F_{1} = \frac{GMm_{1}}{d^{2}}$ and $F_{2} = \frac{GMm_{2}}{d^{2}}$.
Given that $F_{1} = F_{2}$,we have $\frac{GMm_{1}}{d^{2}} = \frac{GMm_{2}}{d^{2}}$.
By canceling the common terms $\frac{GM}{d^{2}}$ from both sides,we get $m_{1} = m_{2}$.
$(b)$ The acceleration due to gravity $g$ for an object of mass $m$ at the surface of the Earth (radius $R$) is given by $g = \frac{GM}{R^{2}}$.
$(c)$ $G$ is called a universal gravitational constant because its value remains the same throughout the entire universe,regardless of the location or the nature of the interacting bodies.