Gravitation

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-d

Explanation- for small objects  sizes less than 100 m centre of mass very close with the centre of gravity of the body. But when the size of object increases it weight changes and its CM and CG become far from each other.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c

Explanation- a geostationary satellite is having same sense of rotation as that of earth i.e west east direction.

A polar satellite goes around earth's pole in north south direction.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a,c,d

Explanation- F₁=-F₂=- $\frac{r_{12}}{r_{12}^3}G{M}_o^2\left(\frac{m_1{m}_2}{M_o^2}\right)$ ⁿ

r₁₂=r₁-r₂

acceleration due to gravity ,g = F/mass= $\frac{G{M}_o^2({m_1{m}_2)}^n}{r_{12}^2{\left(M_o\right)}^{2n}}*\frac{1}{mass}$

as g is constant hence constant of proportionality will not be constant in kepler's third law.

Hence kepler's law will not be valid.

For negative n, g= $\frac{G{M}_o^2{\left(m_1{m}_2\right)}^{-n}}{r_{12}^2{\left(M_o\right)}^{-2n}}*\frac{1}{mass}$

= $\frac{G{M}_o^{2(1+n)}}{r_{12}^2}\frac{{{(m}_1{m}_2)}^{-n}}{mass}$

M_o>m₁ or m₂ so objects lighter than water will sink

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-c, d

Explanation- Gravitation force F= $\frac{GMm}{r^2}$   where M is mass of sun and m is mass of earth.

When G decrease force will also decrease so according to this it will not in the fixed circular orbits around the sun. as when radius increases the force will decreases and soon earth leave the solar system.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c, d

Explanation-due to the huge amount of charges on the sun and the earth electrostatics force of attraction will be large. Gravitational force is also attractive in nature have both forces will be added and obeys kepler's law also.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c, d

Explanation- G'=10G

Consider the diagram

Force on the objects due to the earth = $\frac{G^{\text{'}}Mm}{R^2}=\frac{10GMm}{R^2}$

= (10g)m= 10mg

Force on the objects due to the sun F= $\frac{GM\text{'}m}{r^2}=\frac{GMm}{10r^2}$

As r>>R  so F will be very small

So effect of sun will be neglected

Now g'=10g

Hence weight of person =10mg

So gravity pull on the person will increase, due to it walking on ground would be more difficult

Critical velocity V_c is proportional to g

So V_{c $\propto g$}

V'>g

V_c'>V_c hence rain will fall faster. To overcome the increased gravitational force of the earth the aeroplane will have to travel much faster.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c

Explanation- as we know F= $\frac{GMm}{a^2}=\frac{m{v}^2}{a}$

V= orbital speed= $\frac{\sqrt[]{Gm}}{a}$

Time period of revolution of planet T = $\frac{2\pi a}{v}=\frac{2\pi a}{\frac{\sqrt[]{Gm}}{a}}$ = $\frac{2\pi a^2}{\sqrt[]{GM}}$

T² $\propto a^4$ ……….1

Hence orbit will be elliptical.

F= $\frac{GM}{a^3}m$ =g'm

It is clear from equation 1 that its path is parabola.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-a, c, d

Explanation- acceleration due to gravity at altitude g'= $\frac{g}{{\left(1+\frac{h}{R}\right)}^2}\approx g\left(1-\frac{2h}{R}\right)$

At depth d, g'=g (1-d/R)

In both cases value of g decreases

But in case of latitude the value of g increases when we increase $\varnothing$

$g_{\varnothing }=g-w^{2}R{cos}^2\varnothing$

also we conclude from the formula that it is independent upon mass.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-c

Explanation- force on B due to A = F_BA= $\frac{G\left(2Mm\right)}{{AB}^2}$

force on B due to C = F_BC= $\frac{G\left(Mm\right)}{{BC}^2}$

BC = 2AB

F_BC= $\frac{G\left(Mm\right)}{(2{AB)}^2}=\frac{G\left(Mm\right)}{4{AB}^2}$ BA

So it move towards BA