Mechanics (Part – 3)

Its value for Earth is 9.8 m/s².

It is represented by "g" ie. g= 9.8 m/s²

W = m × g

Where:

• W = weight of the body (in newtons, N)
• m = mass of the body (in kilograms, kg)
• g = acceleration due to gravity (in m/s²)
  ⇒ on Earth, approximately 9.8 m/s².

So, weight is the force with which gravity pulls a body toward the center of a planet.

Galileo demonstrated through experiments (like rolling balls down inclined planes) and theoretical reasoning that:

• The acceleration due to gravity (g) is constant for all objects in free fall.

• Therefore, in a vacuum, all objects fall with the same acceleration and will hit the ground at the same time if dropped from the same height.

"The rate of change of impulse of an object is proportional to the force exerted on it based on Newton’s Second Law of Motion."

Newton’s Second Law states:

Force (F) is equal to the rate of change of momentum (or impulse) with respect to time.

Mathematically:

F = dp/dt,

where p = momentum = mass × velocity

And since impulse = change in momentum, this law also ties force directly to impulse.

• Mass is the amount of matter in an object — it does not change with location.
• Weight depends on gravity, and the Moon’s gravity is about 1/6th that of Earth.
  ⇒ So, on the Moon, the object weighs less, but it still has the same mass.

If the speed of rotation of the Earth increases, the weight of the body decreases.

Explanation:

• Weight is the force with which the Earth pulls an object toward its center, calculated as:

  Weight=mass×effective gravity

• When the Earth rotates faster, the centrifugal force acting on objects (especially at the equator) increases.

• This centrifugal force acts outward, opposing the pull of gravity.

• As a result, the net (effective) gravitational force decreases, making the object weigh less.

A body in equilibrium experiences zero net force and zero net torque.

This can happen in two cases:

1. Static equilibrium – the body is at rest.
2. Dynamic equilibrium – the body is moving at constant velocity (no acceleration).

The height of a geostationary satellite above the surface of the Earth is approximately:

35,786 kilometers (km)

Explanation:

• A geostationary satellite orbits the Earth once every 24 hours, matching the Earth’s rotation.

• It remains fixed over a single point on the equator.

• To achieve this, it must be placed at a precise altitude where the gravitational force and the centripetal force due to orbital motion balance out.

• This altitude from Earth's center is about 42,164 km.

• Since Earth’s radius is about 6,378 km, the height above the surface is:

42,164 km−6,378 km≈35,786 km=36000km

This is the standard altitude for geostationary satellites.