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The state of weightlessness (zero weight) can be observed in the following situations. 

• When objects fall freely under gravity.
• When a satellite revolves in its orbit around the earth.
• When bodies are at null points in outer space. The zero gravity region is called null point.

Since the escape velocity does not depend upon the angle of projection, hence the escape velocity will be same.

Given; potential energy on the Earth’s surface U_i = – 54 J; mass of the body m = 3 kg
If the body escapes, then its potential energy at infinity U_f = 0
Escape energy = U_f – U_i = 0 – (-54) = 54J
or \(\frac{1}{2} m v_{e}^{2}\) = 54
or v_e = \(\sqrt{\frac{2 \times 54}{m}}=\sqrt{\frac{2 \times 54}{3}}=\sqrt{36}\)
or v_e = 6 ms^-1

Orbital velocity of the satellite = v₀
When any hour, its orbital velocity is increases by 41.4% then the new value of orbital velocity
v’₀ = v₀ + \(\frac{41.4}{100} v_{0}=v_{0}\left[1+\frac{41.4}{100}\right]\)
= v₀ [1 + 0.414]
= v₀ × 1.414 = v₀\(\sqrt{2}\)
or v’₀ = v_e (escape velocity)
Therefore, the satellite will escape.

Tidal effect ∝ (1)/(distance)³ whereas gravitational force ∝ (1)/(distance)²

Although sun's pull is more than moon, yet tidal effect due to moon is more because moon is nearer to the earth.

The tidal effect is inversely proportional to the cube of the distance while the gravitational force is inversely proportional to the square of the distance. Since the distance between sun and earth is much greater than that between the moon and earth, the effect observed due to the moon is more that due to the sun.

Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

Newton's law of universal gravitation:-

"According to this law each particle in the universe attracts every other particle in the universe. The force of attraction between them is directly proportional to the square of the distance between them".

Consider two particles of masses m₁ and m₂ separated by a distance r. The force of attraction between them:

F ∝ m₁m₂/r²

or, F = Gm₁m₂/r²

Where G = 6.67 x 10^-11 Nm² kg^-2, is the universal gravitational constant.

Universal gravitational constant:

We have, F = Gm₁m₂/r²

If m₁ = m₂ 1 kg, r = 1 m, then G = F. Thus, universal gravitational constant (G) is numerically equal to the force of attraction between two bodies of mass 1 kg each, separated by a distance of 1 m.

The importance of Newton’s universal law of gravitation :

This law explains successfully, i.e., with great accuracy,

• The force that binds the objects on the earth to the earth
• The motion of the moon and artificial satellites around the earth
• The motion of the planets, asteroids, comets, etc., around the Sun
• The tides of the sea due to the moon and the Sun.

The universal law of gravitation states that every body in the universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

(D) it is applicable at all places of universe for all distances between all particles.

Gravitational constant is numerically equal to the force of attraction between two bodies of unit masses separated by unit distance.

Dimensional formula of G is [M^-1L³ T^-2 ].

The gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. If two masses m₁ and m₂ are separated by a distance r then F ∝ \(\frac {m_{m2}}{r2}\) Or \(F =G\frac {m_1m_2}{r^2}\) Where G is the proportionality constant known as universal gravitational constant.

From the above equation, G = \(\frac {Fr^2}{m_1m_2}\)

If m₁ = m₂ = 1 and r = 1 then G = \(\frac {F(r^2)}{m_1m_2}\)

Gravitational constant is defined as the gravitational force attraction between two bodies of unit masses separated by unit distance. Dimensional formula for [G] = [M^-1 L³ T^-2 ].

Intensity of gravitational field,
E_g = \(\frac{F}{m}\)
Dimensional formula of E_g = \(\frac{\left[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}\right]}{\left[\mathrm{M}^{1}\right]}\)
= [M⁰LT^-2].