All questions of Gravitation for UPSC CSE Exam

Understanding Weight on the Moon
The weight of an object is determined by the formula:
Weight = Mass × Gravitational Acceleration
Here, we can analyze the weight of a 100 kg body on the Moon's surface.
Given Values
- Mass of the body = 100 kg
- Gravitational acceleration on the Moon = 1.6 m/s²
Calculating the Weight
To find the weight of the body on the Moon, we can substitute the values into the formula:
- Weight = 100 kg × 1.6 m/s²
Performing the Calculation
- Weight = 160 N
Thus, the weight of the 100 kg body on the Moon's surface is 160 Newtons.
Conclusion
Therefore, the correct answer is option 'D' (160 N).
- The gravitational force on the Moon is significantly less than on Earth (approximately 9.8 m/s²).
- This lower gravitational acceleration results in a reduced weight for objects on the Moon compared to their weight on Earth.
This concept is essential for understanding how gravity affects objects differently on various celestial bodies.

Understanding Gravitational Force
Gravitational force is a fundamental interaction that occurs between any two objects with mass, and it plays a crucial role in the dynamics of celestial bodies as well as everyday objects.
Characteristics of Gravitational Force
- Universal Attraction: Unlike other forces, gravitational force is always attractive. This means that two masses will always pull towards each other regardless of the distance separating them.
- Distance Independence: The force acts at all distances, meaning that even at vast distances, the gravitational force is present, albeit weaker.
- Inverse Square Law: The strength of the gravitational force decreases with the square of the distance between the two masses. This law means that as you double the distance, the gravitational force becomes four times weaker, but it never becomes zero.
Implications of Gravitational Force
- Cosmic Scale: On a cosmic scale, gravitational force governs the motions of planets, stars, and galaxies, keeping them in orbit and influencing their behaviors.
- Everyday Life: In everyday situations, gravity ensures that objects fall towards the Earth, providing stability and structure to our environment.
- No Repulsive Force: Unlike electromagnetic forces, which can be either attractive or repulsive, gravitational force lacks any repulsive component. This is why the correct answer is "attractive at all distances."
Conclusion
In summary, gravitational force is an ever-present attractive force between masses, operating at all distances without exception. This fundamental nature of gravity is why option 'C' is correct. Understanding this force is essential for comprehending both terrestrial and astronomical phenomena.

Understanding the Universal Gravitational Constant (G)
The Universal Gravitational Constant (G) is a fundamental constant in physics, crucial for understanding gravitational forces between masses. Its value is typically expressed in different units depending on the system used.
Value of G in C.G.S Units
In the C.G.S (centimeter-gram-second) system, the value of the Universal Gravitational Constant is represented as:
- G = 6.67 x 10^-8 cm³/g·s²
This value is derived from the standard value of G in SI units, which is approximately 6.674 x 10^-11 N·m²/kg². To convert it into the C.G.S system, we apply the necessary conversions:
- 1 N (Newton) = 10^5 dyne
- 1 m = 100 cm
- 1 kg = 1000 g
Using these conversions, the value of G in C.G.S units becomes:
- G = 6.674 x 10^-11 N·m²/kg² = 6.67 x 10^-8 cm³/g·s²
Explanation of Options
Now, let’s analyze the options provided:
- a) 6.67 * 10^-6 cgs unit (Incorrect)
- b) 6.67 * 10^-7 cgs unit (Incorrect)
- c) 6.67 * 10^-8 cgs unit (Correct)
- d) 6.67 * 10^-10 cgs unit (Incorrect)
The correct answer is option 'C' (6.67 * 10^-8 cgs unit), which aligns with the established value of G in the C.G.S system.
Conclusion
Understanding the value of G in different units is essential for solving problems in gravitational physics, especially in fields like astrophysics and engineering. The correct conversion and recognition of this constant are crucial for accurate calculations in these domains.