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Initial speed u = 50 m/s ; g = 10 m/s²    

Maximum height reaches by the body (h) = \(\frac{u^2}{2g}=\frac{50\,\times\,50}{2\,\times\,10}\) = 125 m

Time taken to reach the maximum height (t) = \(\frac{u}{g}=\frac{50}{10}\) = 5 sec.

After reaching maximum height, the velocity becomes ‘zero’.

• When the right leg and right shoulder are in contact with the surface of the wall along the height of a man, his weight is towards the wall. 
• The centre of gravity will be away from the foot. Just like in the case of carrying a bucket full of water with one hand. 
• The line joins the centre of gravity and centre of equilibrium is not perpendicular to the horizontal. 
• Hence he cannot lift his left leg without moving his body along the wall.

From Einstein’s special theory of relativity, we know that the speed of light, c = (3 x 10⁸ ms⁻¹ ) cannot be exceeded by any moving particle. Thus c is the upper limit to the projectile escape speed.

i.e., v_e = \(\big(\frac{2GM}{R}\big)^{\frac{1}{2}}\leq e\)

if M = M_e

then R = \(\frac{2GMe}{c^2}\) 

= \(\frac{2\times(6.67\times10^{-11})\times(6\times10^{24)}}{(3\times10^8)^2}\)

= 8.9 x 10^-3 m = 1 cm

It means the earth can act as a black hole. If it shrinks to a very small size of nearby 1 cm radius, which is the size of a berry.

The unit of G is Nm²kg^-2 

Acceleration due to gravity is independent of the mass of the body.

Acceleration due to gravity will decreases if angular speed of rotation of earth increases.

Using F = \(\frac{GMm}{R^2}\)

we have new weight

given by F' = \(\frac{GMm}{(2R)^2}\) = \(\frac{1}{4}\)F

\(\therefore\) Weight becomes one-fourth of the original value.

The value of gravitational potential energy is negative and it increases as we move away from the earth and becomes maximum ( zero) at infinity.

U = GMm/r = (GM/r²) r m = g r m =(mg) r

The relations for orbital velocity is

v_o = \(\sqrt{\frac{Gm}{(R+h)}}\)

Where h is the height of the satellite above the earth’s surface. Clearly, the smaller is the value of h, greater is the value of v and vice-versa. Hence, satellite revolving close to earth has larger speed.

Time period of satellite is given by; T=2π√(R+h)³/GM  Therefore ,T will decrease, when h decreases.

As, escape speed = √(2GM/R), therefore its value are different for different planets which are of different masses and different sizes.

Correct option is: (C) \(m^0\)

Escape velocity \(ve = \sqrt{2gRe}\)

Escape velocity does not depend on the mass.

Correct option is: (C) m⁰

We know that,

(Time period)² \(\propto\) (Orbit radius)³ .

Therefore, time period is larger in case of satellite revolving away from the surface of earth.

We require the same velocity for the two balls, while projecting them out of the gravitational field. It is so because, the value of escape velocity does not depend upon the mass of the body to be projected [i.e. v_e = √(2gR)].

Let v₀,v_e be the orbital and escape speeds of the satellite, then v_e = √2v₀,

Energy in the given orbit, E₁ = 1/2 mv₀² = E

Energy for the escape speed, E₂ = 1/2 m v_e²=1/2m(√2v₀²)=2E

Energy required to be supplied = E₂-E₁ =E.

This can be possible only if we can ignore air friction and technical difficulties.

The true weight of a body will be zero where the gravitational effects are nil, e.g., at the centre of earth.

At perihelion because the earth has to cover greater linear distance to keep the areal velocity constant.

The satellite at the smaller height would have greater velocity. This is because v_o ∝ \(\frac{1}{\sqrt{r}}\).

We will give then the same velocity as escape velocity is independent of the mass of the body.

The angle between the equatorial plane and the orbital plane of

(a) 90^o 

(b) 0^o 

No, the weight of the body is only due to the earth’s gravity.