Gravitation

Statement I is true. Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between their centers. This law applies to any pair of bodies in the universe.

Statement II is also true. The weight of an object is the force of gravity acting on it. If a person were at the center of the Earth, the gravitational pull from all the surrounding mass of the Earth would cancel out, resulting in zero net gravitational force and therefore zero weight.

Therefore, Option A: Both Statement I and Statement II are true, is the correct answer.

Assertion A is false. While the direction of acceleration due to gravity does always point towards the center of the Earth, its magnitude actually varies from the poles to the equator. This is due to the Earth's rotation and the fact that Earth is not a perfect sphere but an oblate spheroid, meaning it's slightly flattened at the poles and bulging at the equator.

Reason R is true. At the equator, the direction of acceleration due to gravity is towards the center of the Earth.

The acceleration due to gravity (g) at a distance (r) from the center of a planet is given by:

$g = \frac{G M}{r^2}$

where:

• G is the gravitational constant,
• M is the mass of the planet,
• r is the distance from the center of the planet.

If the height h of a point P above the surface of the Earth is equal to the diameter of the Earth, then the distance r from the center of the Earth to the point P is 3 times the radius of the Earth. Substituting this into the equation for g gives:

$g' = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$

where:

• g' is the acceleration due to gravity at the point P,
• R is the radius of the Earth.

Therefore, the value of acceleration due to gravity at point P is g/9.

We know that

T² $\propto$ R³

$ \Rightarrow {\left( {{{T'} \over T}} \right)^2} = {\left( {{{3R} \over R}} \right)^3}$

$ \Rightarrow {{T'} \over T} = 3\sqrt 3 $

$ \Rightarrow T' = 3\sqrt 3 $ years

According to the given information

${{GM} \over {{{(R + h)}^2}}} = {1 \over 3} \times {{GM} \over {{R^2}}}$

$ \Rightarrow R + h = \sqrt 3 R$

$ \Rightarrow h = (\sqrt 3 - 1)R \simeq 4685$ km

Inside a hollow spherical shell, net gravitational field is zero at all the points. Hence, force of attraction on a point mass inside shell is always zero. Spherical objects like shells acts like a point mass at centre. This is shell theorem. So statement II is correct.

Escape speed is $v_0$,

$ \Rightarrow \quad v_0=\sqrt{\frac{2 G M}{R}} $

Now, if velocity of object at infinity is $v$, then by conservation of energy we have,

Energy at surface $=$ Energy at infinity

$ \frac{-G M m}{R}+\frac{1}{2} m\left(5 v_0\right)^2=\frac{1}{2} m v^2+0 $

Now, $\quad \frac{G M}{R}=\frac{1}{2} v_0^2$

[from Eq. (i)]

So, $-\frac{1}{2} m v_0^2+\frac{1}{2} m\left(5 v_0\right)^2=\frac{1}{2} m v^2$

$ \begin{array}{rlrl} \Rightarrow &\ -v_0^2+25 v_0^2 & =v^2 \\ \text { or } & & v & =\sqrt{24\left(v_0^2\right)} \\ \text { or } & & v & =2 \sqrt{6} v_0 \end{array} $

The given situation is shown below.

$ A C=B D=\sqrt{\left(4 l_0\right)^2+\left(3 l_0\right)^2}=5 l_0 $

∴ Potential energy of the system,

$ \begin{aligned} U=U_{A B}+U_{B C}+U_{C D} & +U_{D A} +U_{A C}+U_{B D} \end{aligned} $

Applying the formula,

$U=-G \frac{m_1 m_2}{r}$, we get

$ \begin{aligned} U & =-\frac{G \cdot m \cdot m}{4 l_0}-\frac{G \cdot m \cdot m}{3 l_0}-\frac{G \cdot m \cdot m}{4 l_0} \\ & \quad-\frac{G \cdot m \cdot m}{3 l_0}-\frac{G \cdot m \cdot m}{5 l_0}-\frac{G \cdot m \cdot m}{5 l_0} \\ & =-\frac{G m^2}{l_0}\left[\frac{1}{4}+\frac{1}{3}+\frac{1}{4}+\frac{1}{3}+\frac{1}{5}+\frac{1}{5}\right] \\ & =-\frac{G m^2}{l_0}\left[\frac{1}{2}+\frac{2}{3}+\frac{2}{5}\right] \\ & =-\frac{G m^2}{l_0}\left[\frac{15+20+12}{30}\right] \\ & =-\frac{G m^2}{l_0}\left(\frac{47}{30}\right) \\ |U| & =\frac{47}{30}\left(\frac{G m^2}{l_0}\right) \end{aligned} $

The force of attraction between the complete sphere of mass $M$ and particle of mass $m$ is

$ F=\frac{G m M}{4 R^2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)$

Mass of the complete sphere,

$ M=\frac{4}{3} \pi R^3 \rho $

where, $\rho$ is density of sphere. Mass of the cut out portion,

$ m^{\prime}=\frac{4}{3} \pi\left(\frac{R}{2}\right)^3 \rho=\frac{M}{8} $

Now, the distance between the centre of the cut out portion and mass $m$,

$ =2 R-\frac{R}{2}=\frac{3 R}{2} $

Hence, the force of attraction between the cut out portion and mass $m$ is

$ \begin{aligned} F^{\prime} & =\frac{G m^{\prime} M}{\left(\frac{3 R}{2}\right)^2}=\frac{G\left(\frac{M}{8}\right) m}{\frac{9 R^2}{4}} \\ & =\frac{G m M}{4 R^2} \times \frac{2}{9}=\frac{2}{9} F[\text { from Eq }(\text { i) }] \end{aligned} $

Thus, the force of attraction between the remaining part of the sphere and

$ \text { mass } m=F-F^{\prime}=F-\frac{2}{9} F=\frac{7}{9} F $

The given situation is shown below.

When the rocket will reach at maximum height, then the total gravitational potential energy at this height will be equal to the sum of kinetic energy at the surface of earth and the potential energy when the rocket was at the earth's surface.

$ \begin{array}{cc} \therefore & -\frac{G m M_c}{(R+h)}=\frac{1}{2} m v^2+\left[\frac{-G m M_c}{R}\right] \\ \Rightarrow & \frac{-G m M_c}{(R+h)}=\frac{1}{2} m v^2-\frac{G m M_c}{R} \\ \Rightarrow & \frac{1}{2} m v^2=\frac{G m M_c}{R}-\frac{G m M_c}{R+h} \\ \Rightarrow & \frac{1}{2} v^2=G M_c\left[\frac{1}{R}-\frac{1}{R+h}\right] \\ \Rightarrow & \frac{1}{2} \times(4000)^2=\frac{G M_c}{R^2} \times R^2\left[\frac{R+h-R}{R(R+h)}\right] \\ \Rightarrow & 8 \times 10^6=g R^2\left[\frac{h}{R(R+h)}\right] \\ & =\frac{g h R}{(R+h)}=\frac{10 h R}{(R+h)} \end{array} $

$ \begin{aligned} & \Rightarrow \quad 8 \times 10^5=\frac{h R}{R+h} \\ & \Rightarrow 8 \times 10^5(R+h)=h R \\ & \Rightarrow 8 \times 10^5\left[6.4 \times 10^6+h\right]=h \times 6.4 \times 10^6 \\ & \Rightarrow 51.2 \times 10^{11}=\left(6.4 \times 10^6-8 \times 10^5\right) h \\ & \Rightarrow 51.2 \times 10^{11}=\left(64 \times 10^5-8 \times 10^5\right) h \\ & \Rightarrow \quad h=\frac{51.2 \times 10^{11}}{56 \times 10^5}=0.914 \times 10^6 \mathrm{~m} \\ & \Rightarrow \quad h=\frac{0.914 \times 10^6}{10^3} \mathrm{~km}=914 \mathrm{~km} \end{aligned} $

The situation given in the question is depicted below

Let the speed of particle is $v$ for circular motion.

Clearly the gravitational force acting all the 3 masses will be same due to symmetry and let it be $F$.

Now, considering the mass $M$ placed at point $B$. Since, it is performing circular motion in radius say $R$, hence the net gravitational pull on it due to the other two masses are providing the required centripetal force.

Now, net force on $M$ at $B$,

$ \begin{aligned} & F^{\prime} =\sqrt{F^2+F^2+2 F^2 \cos 60^{\circ}} \\ \Rightarrow F^{\prime} & =\sqrt{3} F \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } & F =\frac{G M^2}{l^2} \end{aligned} $

(gravitational force of Newton)

Also, by symmetry of forces the $F^{\prime}$ will be acting along the bisector of angle at point $B$ i.e. at $30^{\circ}$ from each force $F$.

Now, in $\triangle O B P$,

$ \begin{array}{rlrl} \cos 30^{\circ} & =\frac{\frac{l}{2}}{R} \left(\because B P=\frac{B C}{2}\right) \\ \frac{\sqrt{3}}{2} & =\frac{l}{2 R} \\ \Rightarrow \quad R & =\frac{l}{\sqrt{3}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

Now, since $F^{\prime}=F_{\text {centripetal }}$

i.e. $\quad F^{\prime}=F_c$

$ \begin{array}{ll} \Rightarrow \quad \sqrt{3} F=\frac{M v^2}{R} & \text { [from Eq. (i)] } \\ & {\left[\because F_c=\frac{M v^2}{R}\right]} \end{array} $

$ \Rightarrow \quad \sqrt{3} \frac{G M^2}{l^2}=\frac{M v^2}{R} $

$ \begin{aligned} & \sqrt{3} \frac{G M^2}{l^2} =\frac{M v^2}{l} \sqrt{3} \\ \Rightarrow & v^2 =\frac{G M}{l} \\ \Rightarrow & v =\sqrt{\frac{G M}{l}} \end{aligned} $

Time period of satellite moving around a planet with angular velocity $\omega$ is given as,

$T=\frac{2 \pi}{\omega}$

Since, the two artificial satellites revolving in the same circular orbit, then their angular velocities will be same.

Hence, both satellites revolve with same period of revolution.

Orbital velocity of revolving satellite, $v=\sqrt{\frac{G m}{r}}$ i.e. $v \propto \frac{1}{\sqrt{r}}$

Where, $m=$ mass of the planet

$G=$ Universal gravitational constant

$r=$ radius of circular orbit.

Escape velocity depends on the gravitational potential at the point from where the body is launched. Since, this potential depends on the height of the point of projection, hence escape velocity also depends on the height (altitude) of the point of projection.

Since, uniform solid sphere of radius $R$ produces a gravitational acceleration $a_0$ on its surface, hence $a_0=\frac{G m}{R^2}$

where, $m$ is the mass of solid sphere.

Let $x$ be the distance of the point from the surface of sphere where the gravitational acceleration becomes $\frac{a_0}{4}$.

i. e. $\frac{a_0}{4}=\frac{a_0}{\left(1+\frac{x}{R}\right)^2} \Rightarrow\left(\frac{1}{2}\right)^2=\frac{1}{\left(1+\frac{x}{R}\right)^2}$

$\begin{aligned} \Rightarrow & & \frac{1}{2} & =\frac{1}{1+\frac{x}{R}} \Rightarrow 1+\frac{x}{R}=2 \\ \Rightarrow & & x & =R \end{aligned}$

Hence, distance of the point from the centre of sphere $=x+R=R+R=2 R$

Initial speed of projectile from the surface of earth, $v=\alpha v_{\text {e }}$

Where, $\alpha$ is constant and $v_e$ is escape velocity of projectile.

Radius of earth, $R_E=6400 \mathrm{~km}$

$\qquad=6.4 \times 10^6 \mathrm{~m}$

Maximum height attained by the projectile,

$h=800 \mathrm{~km}=8 \times 10^5 \mathrm{~m}$

We know that, escape velocity on the surface of earth, $v_e=11.2 \mathrm{~km} / \mathrm{s}$

$\begin{aligned} & =11.2 \times 10^3 \mathrm{~m} / \mathrm{s} \\ v & =\alpha v_e \\ v & =\alpha \times 11.2 \times 10^3 \quad \text{... (i)} \end{aligned}$

According to given situation, applying the law of conservation of energy

$\begin{aligned} & \frac{1}{2} m v^2=m g h \\ \Rightarrow & v^2=2 g h \Rightarrow\left(\alpha v_e\right)^2=2 g h \\ \Rightarrow \quad & \alpha=\frac{\sqrt{2 g h}}{v_e}=\frac{\sqrt{2 \times 10 \times 8 \times 10^5}}{11.2 \times 10^3}=\frac{4}{11.2} \approx \frac{1}{3} \end{aligned}$

Let us consider the gravitational force acting on each mass M by adjacent particles be F.

and the gravitational force acting on each mass M diagonally be F₁.

Along the centre of circle,

The problem describes two identical particles of mass m=1kg each moving in a circle of radius R under the action of their mutual gravitational attraction. This means that the particles are moving around a common center, and the distance between the particles is 2R (as the diameter of the circle).

In this case, the force providing the centripetal force for each particle to move in a circular path is the gravitational force between the particles.

The gravitational force between two masses m1 and m2 separated by a distance r is given by Newton's law of universal gravitation:

$F = G \frac{{m_1 m_2}}{{r^2}}$

Since the two particles are identical, m1=m2=m=1kg. And the distance between them r is 2R. Substituting these into the above equation gives the gravitational force between the two particles:

$F = G \frac{{m^2}}{{(2R)^2}} = G \frac{{1}}{{4R^2}}$

This gravitational force is also equal to the centripetal force needed for each particle to move in a circular path of radius R. The centripetal force is given by:

$F = m R ω^2$

Setting these two equations equal to each other gives:

$G \frac{{1}}{{4R^2}} = 1 \cdot R \cdot ω^2$

Rearranging this to solve for ω (the angular speed) gives:

$ω = \frac{1}{2} \sqrt{\frac{G}{R^3}}$

So, the angular speed of each particle is: $\frac{1}{2} \sqrt{\frac{G}{R^3}}$.

In a binary star system, the two stars orbit around a common center of mass. When considering periods of revolution, Kepler's Third Law comes into play. This law states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (r) of the orbit. It's often written in the following form for a single object orbiting another:

T² ∝ r³

For a binary star system, this would still hold true. The periods of revolution for both stars A and B will be the same because they are both orbiting the same common center of mass, regardless of their individual masses or individual orbital radii. In other words, star A and star B complete one orbit in the same amount of time.

So, Option B: T_A = T_B is correct.

The other options (Option A, C, and D) would not be correct. Kepler's Third Law is not a ratio between the periods and radii of two different bodies, and the periods do not depend on the masses of the individual stars or their individual distances from the center of mass.

Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:

$T^2 \propto r^3$

where:

• $T$ is the orbital period
• $r$ is the radius of the circular orbit (which serves as the semi-major axis in this case)

Considering two satellites, one with period $T$ and radius $R$, and another with unknown period $T'$ and radius $9R$, we can form an equation:

$\frac{{T'}^2}{T^2} = \frac{(9R)^3}{R^3}$

This simplifies to:

$\frac{{T'}^2}{T^2} = 729$

Taking the square root of both sides, we get:

$T' = T \times \sqrt{729} = T \times 27$

So, the period of another satellite in a circular orbit of radius $9R$ is $27T$.

According to Kepler’s second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.