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For a body to be a black hole, even light should not escape. So limiting escape velocity is 3 × 10⁸ ms^-1 

So, for body of Mass M, condition is,

\(\sqrt \frac {2GM}{R}\) ≥ 3 × 10⁸ ms^-1

For Earth, M = 6 × 10²⁴ kg

\(\frac {2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{R}\) = (3 × 10⁸)² R

= 9 × 10^-2 m

R = 9 cm.

We are given that: Initial velocity (u) = 20 m/s

When an object attain maximum height, then final velocity (ν) = 0

a = g = 10m/s², time (t) = ?

(1) From first equation of motion, we have v = u – gt

⇒ 0 = 20 - 10 x t 

⇒ 10t = 20 

⇒ t = 2 sec

(2)  From third equation of motion, we have
v² = u² – 2gh

⇒ (0)² = (20)² - 2 x 10 x h

⇒ 20h = 20 x 20

⇒ h = 20 m

We are given that: Height of tower (h) = 125 m

Initial velocity (u) = 0

g = 10 m/s²

(1) From 2nd equation of motion

For downward motion: h = ut + \(\frac{1}{2}\)gt²

⇒ 125 = 0 x t + \(\frac{1}{2}\) x 10 x t² = 125 = 5t² 

⇒ t² = 25 

⇒ t = 5sec²

(2) From 1st equation of motion,

v = u + gt

⇒ u = 0 + 10 x 5

⇒ v = 50 m/s

We know that the value of g is inversely proportional to the square of radius or g ∝ \(\frac{1}{R^2}\).

Radius of earth is minimum at poles, so acceleration due to gravity is maximum at poles. Similarly, radius of earth is maximum at equator, so acceleration due to gravity is minimum at equator. Also, acceleration due to gravity decreases as we go higher from the earth surface.

Hence, option D is correct.

When a ball is thrown vertically upward, then acceleration due to gravity is directed opposite to the direction of motion since the ball experiences a force of attraction directed towards the center of earth due to earth’s gravity.

Direction of acceleration is in the direction of force as,

a = \(\frac{F}{m}\)

So, from above, we can conclude that acceleration is also directed towards the centre of earth which is opposite to the vertically upward motion of the ball.

Hence, option A is correct.

Here,

u = Initial velocity = 0

v = Final velocity

s = Distance travelled = 20 m

a = acceleration = g = 10 ms-2

Now, we know from the equations of motion that

v² = u² + 2as

v = \(\sqrt {u^2\,+\,2as}\)

v = \(\sqrt {0\,+\,2\,*\,10\,*\,20}\)

v = \(\sqrt {400}ms^{-1}\)

v = 20 ms^-1

Hence, option D is correct.

The answer is (C) – 20 m/s

The coin reaches the ground first as compared to the piece of paper because it experiences lesser resistance from air than that felt by paper. If the coin and the piece of paper are dropped in vacuum, both of them will touch the ground at the same time.

No. As we go above the surface of the earth, our weight will go on decreasing.

The energy stored in a body due to the gravitational force between the body and the earth is called the gravitational potential energy.

Gravitational potential energy of a body of mass m = \(-\frac{GMm}{R+h}\), where G = gravitational constant, M = mass of the earth, R = radius of the earth, h = height of the body from the surface of the earth.

[Note : As the body is bound to the earth due to the earth’s gravitational froce, the gravitational potential energy of the body is negative. If the body is given kinetic energy equal to \(\frac{GMm}{R+h}\) the body will overcome the earth’s gravitational force. It will then move to infinity and come to rest there.]

Answer is No.

Answer is Zero.

When a body is thrown vertically upward from the surface of the earth, the minimum initial velocity of the body for which the body is able to overcome the downward pull by the earth and can escape the earth forever is called the escape velocity.

In general, when a body is thrown vertically upward from the earth’s surface, its velocity goes on decreasing and after some time the body falls back to the ground. If its initial velocity is increased, the maximum height attained by it is more, but it does fall back to the ground. If the initial velocity is increased continuously, for a particular initial velocity, the body can overcome the earth’s gravitational force and move to infinity and come to rest there. This velocity is called the escape velocity.

Gravitational potential at point is defined as the work done in bringing a body of unit mass from infinity to that point. Its unit is J/kg, denoted by V Gravitational field intensity is defined as the force experienced by a unit mass at that point. It is expressed in N/kg, denoted by I.

l = -v/R.

We know that, T² ∝ R³

\((\frac{T_1}{T_2})^2 = \) \((\frac {R_1}{R_2})^3\)

T₁ = (1/2)^3/2T

= \(\frac {T}{2\sqrt2}\) 

T₁ = 129 days

The year of the new Earth will have 236 days lesser.

Consider M to be mass of Earth and R to be radius of Earth. 

1. At a point ‘h’ above the ground

New g is g_n = \(\frac {GM} {(R+h)^2}\)

= \(\frac {GM} {R^2[1+\frac{h}{R}]^2}\)

g_n = g\([1+\frac {h}{R}]^{-2}\)

if h << R,

g_n = g\([1-\frac {2h}{R}]\)

So, g decreases with increase in height,

2. At a depth ‘d’ below the ground. The new g, gd is given by

d_d = g\([1-\frac {d}{R}]\)

So, g also decreases with depth.

Consider a body of mass m is rotating around the star s in circular path of radius r.

(i) Orbital velocity –

vo = \(\sqrt{\frac{GM}{r}}\) or v ∝ \(\frac{1}{\sqrt r}\)

Orbital Velocity decreases

(ii) Angular velocity \(\frac{2\pi}{T}\)

By Kepler’s III law

T² \(\propto\) r³ or T² = Kr³

\(\omega\) = \(\frac{2\pi}{Kr^{\frac{3}{2}}}\) or \(\omega\) \(\propto\) \(\frac{1}{\sqrt{r^3}}\)

Hence, angular velocity decreases.

(iii) Kinetic Energy, K = \(\frac{1}{2}\)m\(\frac{GM}{r}\) or K \(\propto\) \(\frac{1}{r}\)

Hence K, decreases on increasing the radius.

(iv) Gravitational Potential Energy,

U = \(-\frac{GMm}{r}\)

or U \(\propto\) \(\frac{-1}{2}\)

So, on increasing radius of circular orbit the U increases.

(v) Total energy,

E = K + U = \(\frac{GMm}{2r}+\big(-\frac{Gmm}{r}\big)\)

E = −\(\frac{GMm}{2r}\)

So, increasing the radius, E will also be increased.

(vi) Angular momentum, L = mvr = mr\(\sqrt{\frac{GM}{r}}\)

L = m\(\sqrt{GMr}\) or L \(\propto\) \(\sqrt r\), increases

Let us consider the following diagram in which size point masses are placed at six sides.

A, B, C, D, E, F.

AC = AG + GC + = 2AG

= 2I cos 30º

= 2I\(\sqrt{\frac{3}{2}}\) = \(\sqrt3\) I = AE

AD = AH + HJ + JD

= I sin 30º + I + I sin 30º

= 2 I

Force on A due to B, F₁ = \(\frac{GMm}{I^2}\) along A to B

= \(\frac{Gm^2}{I^2}\)

Force on A due to C,F₂ = \(\frac{Gm.m}{(\sqrt3I)^2}\)

= \(\frac{Gm^2}{3I^2}\)  along A to C [\(\because\) AC = \(\sqrt3I\)]

Force on A due to D,F₃ = \(\frac{Gm.m}{(\sqrt2I)^3}\)

= \(\frac{Gm^2}{4I^2}\) along A to D [\(\because\) AD = 2I]

Force on A due to E,, F₄ = \(\frac{Gm.m}{(\sqrt3I)^2}\)

= \(\frac{Gm^2}{3I^2}\) along A to E

Force on A due to F, F₅ = \(\frac{Gm.m}{(I)^2}\) along A to F

= \(\frac{Gm^2}{I^2}\) along A to F

Resultant force due to F₁ and F₅,

F₁ = \(\sqrt{F_1^2+F^2_5+2F_2F_5\cos120^o}\)

= \(\frac{Gm^2}{I^2}\) along A to D [Angle b/w F₁ and F₅= 120º]

Resultant force due to F₂ and F₄,

F₁ = \(\sqrt{F_1^2+F^2_4+2F_2F_4\cos60^o}\)

= \(\frac{\sqrt3Gm^2}{3I^2}\) = \(\frac{Gm^2}{\sqrt3I^2}\) along A to D

\(\because\) Net force along A to D = F₁ + F₂ + F₃

= \(\frac{Gm^2}{I^2}+\frac{Gm^2}{\sqrt3I^2}+\frac{Gm^2}{4I^2}\)

= \(\frac{Gm^2}{I^2}\)\(\big(1+\frac{1}{\sqrt3}+\frac{1}{4}\big)\)

(a) Angle between equatorial plane and orbital plane is 90º . 

(b) Angle between equatorial and orbital plane is 0º .