Practice free Gravitation (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
In the provided text, what are the units given for the universal gravitational constant G?
Problems 7.17 and 7.19 consistently mention 'G = $6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$'. This translates to $N \text{ m}^2 / \text{ kg}^2$.
Which of the following quantities is NOT conserved when considering the motion of an object under the gravitational influence of another object, according to the 'Points to Ponder'?
Points to Ponder, point 1, states: 'In considering motion of an object under the gravitational influence of another object the following quantities are conserved: (a) Angular momentum (b) Total mechanical energy. Linear momentum is not conserved'.
An astronaut experiences weightlessness in a space satellite primarily because:
Points to Ponder, point 4, clearly states: 'An astronaut experiences weightlessness in a space satellite. This is not because the gravitational force is small at that location in space. It is because both the astronaut and the satellite are in 'free fall' towards the Earth.'
What are the standard units of the universal gravitational constant G, as provided in the NCERT text?
The NCERT text, specifically in problem 7.17 and 7.19, explicitly states G = $6.67 \times 10^{-11} \ N \ m^2 \ kg^{-2}$. This indicates the units are Newtons times meters squared per kilogram squared.
When considering the gravitational force, the universal gravitational constant 'G' is derived from which law of gravitation?
Though not explicitly stated in the provided excerpts, the constant G is intrinsic to Newton's Law of Universal Gravitation, which underpins all the calculations and concepts discussed in the 'Gravitation' chapter. The chapter refers to the gravitational force formula $F = G M_E m / R_E^2$ repeatedly, indicating its origin.
According to the NCERT text, the measurement of the universal gravitational constant G, combined with knowledge of 'g' (acceleration due to gravity) and the Earth's radius (R_E), allows for the estimation of which physical quantity?
The text states: 'The measurement of G by Cavendish’s experiment (or otherwise), combined with knowledge of g and R_E enables one to estimate $M_E$ from Eq. (7.12). This is the reason why there is a popular statement regarding Cavendish : “Cavendish weighed the earthâ€.' (Page 133, Section 7.5)
What is the approximate numerical value of the universal gravitational constant (G) mentioned in the provided numerical problems?
Problems 7.17 and 7.19 explicitly provide the value of G as $6.67 \times 10^{-11} \ N \ m^2 \ kg^{-2}$.
If the universal gravitational constant (G) were to suddenly increase, what would be the immediate effect on the gravitational force between two masses?
The gravitational force is directly proportional to the gravitational constant G, as indicated by the formula $F = G M_1 M_2 / r^2$. Therefore, an increase in G would lead to an increase in the gravitational force between any two masses.
Which of the following statements about the universal gravitational constant (G) is correct?
The term 'universal gravitational constant' itself implies that G is constant throughout the universe and does not depend on the specific masses or the medium. The text uses a single value for G in all calculations, reinforcing its universal nature.
The formula for acceleration due to gravity on the Earth's surface, $g = GM_E / R_E^2$, clearly shows the relationship between 'g', the Earth's mass ($M_E$), its radius ($R_E$), and the universal gravitational constant (G). This equation implicitly suggests that:
From Eq. (7.12), $g = GM_E / R_E^2$, it is clear that 'g' is directly proportional to both G and $M_E$ and inversely proportional to the square of $R_E$.
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