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Two identical object each of mass 50 kg are kept at a distance of separation of 50 cm apart on a horizontal table. The net gravitational force at the mid point of the line joining their centre is.?
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Two identical object each of mass 50 kg are kept at a distance of sepa...
**Solution:**

Given:
Mass of each object, m = 50 kg
Distance of separation between the objects, d = 50 cm = 0.5 m

We need to find the net gravitational force at the midpoint of the line joining their centers.

**Step 1: Calculating the gravitational force between the objects**
The gravitational force between two objects can be calculated using Newton's law of universal gravitation:

F = (G * m1 * m2) / r^2

Where:
F is the gravitational force between the objects,
G is the gravitational constant (6.67430 × 10^-11 N m^2/kg^2),
m1 and m2 are the masses of the objects, and
r is the separation between the centers of the objects.

In this case, both objects have the same mass (m1 = m2 = 50 kg), so the equation becomes:

F = (G * m^2) / r^2

**Step 2: Calculating the gravitational force at the midpoint**
The net gravitational force at the midpoint of the line joining the centers of the objects is the vector sum of the gravitational forces between each object and the midpoint.

Since the objects are identical and equidistant from the midpoint, the magnitudes of the gravitational forces on the midpoint due to each object will be equal.

Let's consider one of the objects. The distance between the midpoint and the object's center is half the separation distance between the objects, i.e., d/2 = 0.25 m.

Using the formula for gravitational force, we can calculate the magnitude of the gravitational force on the midpoint due to one object:

F1 = (G * m^2) / (d/2)^2

Since there are two identical objects, the total net gravitational force at the midpoint is the sum of the gravitational forces due to each object:

F_net = F1 + F1

F_net = 2 * F1

**Step 3: Evaluating the net gravitational force**
Now, let's substitute the given values into the equation and calculate the net gravitational force:

F_net = 2 * [(6.67430 × 10^-11 N m^2/kg^2) * (50 kg)^2 / (0.25 m)^2]

F_net = 2 * (6.67430 × 10^-11 N m^2/kg^2) * (50 kg)^2 / (0.25 m)^2

F_net = 2 * (6.67430 × 10^-11 N m^2/kg^2) * (50 kg)^2 / (0.0625 m^2)

F_net = 2 * (6.67430 × 10^-11 N m^2/kg^2) * (2500 kg^2) / (0.0625 m^2)

F_net = 2 * (6.67430 × 10^-11 N m^2/kg^2) * (2500 * 10^6 kg^2) / (0.0625 m^2)

F_net = 2 * (6.67430 × 10^-11 N m^2/kg^2) * (2.5 * 10^9 kg^2) / (0.0625 m^2)

F_net = (6.67430 × 10
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Two identical object each of mass 50 kg are kept at a distance of separation of 50 cm apart on a horizontal table. The net gravitational force at the mid point of the line joining their centre is.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two identical object each of mass 50 kg are kept at a distance of separation of 50 cm apart on a horizontal table. The net gravitational force at the mid point of the line joining their centre is.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two identical object each of mass 50 kg are kept at a distance of separation of 50 cm apart on a horizontal table. The net gravitational force at the mid point of the line joining their centre is.?.
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