Tags Gravitation

Gravitation

Get insights from 91 questions on Gravitation, answered by students, alumni, and experts. You may also ask and answer any question you like about Gravitation

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

6 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA: AB will be x : y. The value of x is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$\frac{G M}{{(3 R / 2)}^{2}} = \frac{G M}{R^{3}} * r$

$\Rightarrow O A = \frac{4 R}{9} = r$

$A B = R - \frac{4 R}{9} = \frac{5 R}{9} \Rightarrow O A : A B = 4 : 5 = x : y \Rightarrow x = 4$

0 0

New answer posted

6 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be:

(Radius of earth will, be:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 - 2 * \frac{32}{6400})$

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} * 100 = 1 %$

0 0

New answer posted

6 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

A body is projected vertically upward from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be:

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

$V_{C} = \sqrt[]{\frac{2 G M}{R}}$

Using conservation of Mechanical Energy

$\Rightarrow \frac{- G M m}{R} + \frac{1}{2} * m (\frac{V e^{2}}{9}) = \frac{- G M m}{(R + h)}$

$\Rightarrow \frac{1}{R + h} = \frac{8}{9 R} \Rightarrow h = \frac{R}{8} = \frac{6400}{8} = 800 k m$

0 0

New answer posted

6 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Assume there are two identical simple pendulum clocks. Clock – 1 is placed on the earth and Clock – 2 is placed on a space station located at a height h above the earth surface. Clock – 1 and Clock – 2 operate at time periods 4s and 6s respectively. Then the value of h is – (consider radius of earth R_E = 6400km and g on earth 10 m/s²)

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

$\frac{T_{1}}{T_{2}} = \frac{4}{6} = \sqrt[]{\frac{g_{2}}{g_{1}}}$

$\Rightarrow {(1 + \frac{h}{R})}^{2} = \frac{9}{4} \Rightarrow 1 + \frac{h}{R} = \frac{3}{2} \Rightarrow h = \frac{R}{2} = 3200 k m .$

0 0

New answer posted

7 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

On the $x$ -axis and at a distance $x$ from the origin, the gravitational field due to a mass distribution is given by $\frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$ in the $x$ -direction. The magnitude of gravitational potential on the $x$ -axis at a distance $x$ , taking its value to be zero at infinity is:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

0 0

New answer posted

7 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

$g = \frac{A x}{{(x^{2} + a^{2})}^{3 / 2}}$

$\Rightarrow \int_{v}^{0} ? d V = - \int_{x}^{\infty} ? g d x$

$\Rightarrow O - V = - [\int \frac{A x}{{(a^{2} + x^{2})}^{3 / 2}}]$

Let, $a^{2} + x^{2} = t^{2}$

$\Rightarrow 2 x d x = 2 t d t$

$\Rightarrow x d x = t d t$

$\Rightarrow V = \int \frac{A t d t}{t^{3}} \Rightarrow - \frac{A}{t} \Rightarrow - {\frac{A}{\sqrt[]{a^{2} + x^{2}}}|}_{x}^{\infty}$

$\Rightarrow V = \frac{A}{\sqrt[]{a^{2} + x^{2}}}$

0 0

New answer posted

7 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Two planets A and B of equal mass are having their period of revolutions T_A and T_B such that T_A = 2T_B. These planets are revolving in the circular orbits or radii r_A and r_B respectively. Which out of the following would be the correct relationship of their orbits?

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

According to kepler's third law of time period –

${(\frac{T_{A}}{T_{B}})}^{2} = {(\frac{r_{A}}{r_{B}})}^{3} \Rightarrow \frac{r_{A}}{r_{B}} = 4^{1 / 3}$

$\Rightarrow r_{4}^{3} = 4$ $r$ $B$ $3$

0 0

New answer posted

7 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

A body is moving in a low circular orbit about a planet of mass $M$ and radius $R$ . The radius of the orbit can be taken to be $R$ itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is :

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Sol. $\overset{?}{U} = 5 \hat{j}$

$\overset{?}{a} = 10 \hat{i} + 4 \hat{j}$

$\Rightarrow \overset{?}{S} = \overset{?}{U} t + \frac{1}{2} (\overset{?}{a}) t^{2}$

$\Rightarrow 20 \hat{i} + y_{0} \hat{j} = (5 t^{2}) \hat{i} + (5 t + 2 t^{2}) \hat{j}$

$\begin{array}{r} 20 = 5 t^{2} y_{0} = 5 t + 2 t^{2} \\ t = 2 \Rightarrow 18 m . \end{array}$

0 0

New answer posted

7 months ago

Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is

0 Follower 7 Views

Payal Gupta

Contributor-Level 10

0 0

New answer posted

8 months ago

Gravitation Physics NCERT Exemplar Solutions Class 11th Chapter Eight

The centre of mass of an extended body on the surface of the earth and its centre of gravity

(a) Are always at the same point for any size of the body.

(b) Are always at the same point only for spherical bodies.

(c) Can never be at the same point.

(d) Is close to each other for objects, say of sizes less than 100 m. (e) both can change if the object is taken deep inside the earth.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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.

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Gravitation

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience