Test: The Gravitational Constant & Acceleration due to Gravity of the Earth (August 10) - NEET MCQ

To find the mass of the girl, we can use the formula:
Weight (W) = Mass (m) × Acceleration due to gravity (g)
Given: Weight of the girl (W) = 450 N
Acceleration due to gravity (g) = 9.8 m/s² (approximately)
Using the formula, we can calculate the mass (m):
450 = m × 9.8
m = 450/9.8 = 45.9 kg
So, the mass of the girl is 45.9 kg.

Explanation:

Understanding the Concept:

• The weight of an object is the force of gravity on the object. On the moon, the gravity is only about 1/6th as strong as on the Earth. Therefore, an object will weigh less on the moon than it does on Earth.

Calculation:

• The weight of an object on the moon can be found by multiplying its mass by the gravitational force of the moon. The gravitational force on the moon is approximately 1.6 N/kg. So, if the mass of the object is 12 kg, its weight on the moon would be 12 kg * 1.6 N/kg = 19.2 N.

Answer:

• So, the weight of the object on the moon is approximately 19.2 N. The closest answer is 19.6 N (Option A). Please note that the exact answer may vary depending on the exact value used for the moon's gravitational force.

For more detailed explanations and other educational material, visit EduRev.

The point where the gravitational field equals zero is the point where the gravitational forces due to both bodies cancel each other out. This point is known as the neutral point. At this point, the gravitational potential is given by:

 

Where:

r₁r1​ is the distance from the neutral point to the body with mass mm.
r₂r2​ is the distance from the neutral point to the body with mass 9m9m.
Since the neutral point is equidistant from both bodies, . Let's denote this common distance as 
rr. Therefore, we have:

Solving for rr: 

The gravitational potential at the neutral point is given by:

Substitute the value of rr:


Since division by zero is undefined, the gravitational potential at the neutral point is not defined, and the answer is not meaningful. Therefore, the answer is Option A: −20 GMR.

Concept:

The motion under gravity (free-fall): A body is said to be free-falling when its motion is under the influence of gravity only. It is a straight-line motion with a constant acceleration.

Explanation:

In the case of free falling, the acceleration is constant i.e., a = g = 9.8 m/s2, and directed vertically downwards.
In free falling there is no resistance to the motion, so the force of the medium has no effect on the motion.
Weight = 0 

Explanation of the Problem:

• The escape velocity of a body on the earth's surface is the minimum velocity that a body must have in order to escape the gravitational pull of the earth.
• In this problem, the body is projected upward with a velocity that is twice the escape velocity, hence we need to find the velocity of this body at an infinite distance from the center of the earth.

Formula:

• The formula to calculate the escape velocity is √(2GM/R) where G is the gravitational constant, M is the mass of the earth, and R is the radius of the earth.
• Since this formula is already used to calculate the escape velocity, we can use the same formula to calculate the velocity of the body at an infinite distance.

Calculation:

• If we substitute the given values into the formula, we get the escape velocity as 11.2 km/s.
• When the body is projected upward with a velocity twice the escape velocity (i.e., 22.4 km/s), the velocity of the body at an infinite distance will be:

Velocity at infinite distance = (Initial velocity)^2 - (Escape velocity)^2

• Substituting the given values, we get:

Velocity at infinite distance = (22.4 km/s)^2 - (11.2 km/s)^2

• After calculating the above expression, we get the velocity at infinite distance as 11.22 km/s.

Conclusion:

• Therefore, the velocity of the body at an infinite distance from the center of the earth will be 11.22 km/s. Hence, the correct answer is Option A: 11.22 km/s.

Reference:

• For more information on this topic, you can visit [EduRev](https://www.edurev.in/).

To find the relationship involving the period T of a satellite orbiting just above the Earth's surface, we can use the formula for gravitational force.

The gravitational force acting on the satellite can be expressed as:

• F = G × (m × M) / r²

Here, m is the mass of the satellite, M is the mass of the Earth, r is the radius of the Earth, and G is the universal gravitation constant.

For a satellite in circular motion, this gravitational force provides the necessary centripetal force:

• F = m × (v² / r)

Setting these two expressions for force equal gives:

• G × (m × M) / r² = m × (v² / r)

By simplifying, we can eliminate m:

• G × M / r = v²

Since the period T is related to the velocity v by the equation v = 2πr / T, we can substitute:

• G × M / r = (2πr / T)²

Rearranging this equation gives:

• T² = (4π²r³) / (G × M)

Using the density d of the Earth, we can express M as:

• M = d × (4/3)πr³

Substituting this into the period formula leads to:

• T² = (3π / Gd) × r²

Since we are considering the satellite just above the Earth's surface, the relevant relationship simplifies, and we see that:

• 3π / (G × d) is proportional to T².

Thus, we conclude that the quantity 3π/Gd represents T^2.