JEE Physics Spl ncert solutions physics class 11th Class 11 Exam Science Physics Gravitation

What is the concept of weightlessness in gravitation and when does it occur?

3 Views | Posted 5 months ago

1 Answer

Answered by

Pallavi Pathak | Contributor-Level 10

5 months ago

When there is no normal force acting on a body, weightlessness is experienced, and the body free fall under gravity. The concept of gravity is still acting on the body, however, there is no weight sensation of any reaction force from any surface or ground. The weightlessness felt by the astronauts in an orbiting spacecraft is not because gravity is not present but because while they are moving forward, both, they and the spacecraft are in continuous free fall towards the Earth. The role of the normal force in defining weight can be understood from this phenomenon also how it is different from the actual gravitational force.

0 Upvotes 0 Downvotes

Similar Questions for you

The radii of circular orbits of two satellites A and B of the earth, are 4R and R, respectively. If speed of satellite A is 2V, then the speed of satellite B will be

alok kumar singh

If an object is having same weight at same distance above and below the surface of earth, find its distance from surface of earth.

alok kumar singh

$\frac{G M m}{{(R + x)}^{2}} = \frac{G M m (R - x)}{R^{3}}$

->R³ = (R + x)² (R – x)

->R³ = (R²– x²) (R + x)

->x² + Rx – R² = 0

$x = \frac{- R \pm \sqrt[]{R + 4 R^{2}}}{2}$

$x = \frac{(\sqrt[]{5 R} - 1)}{2} R$

At a point away from planet of radius 6400 km, the gravitational potential and field are –6.4 × 10⁷ SI units and 6.4 SI units respectively. Find height of that point above surface of planet.

alok kumar singh

$\frac{G M}{r} = 6.4 * 1 0^{7}$ . (i)

$\frac{G M}{r^{2}} = 6.4$ . (ii)

$r = \frac{6.4 * 1 0^{7}}{6.4}$

= 10⁷ m

= 10,000 km

R + h = 10,000

h = 10,000 – 6400 = 3600 km

Mass of moon is $\frac{1}{81}$ times the mass of a planet and radius is $\frac{1}{9}$ times the radius of the planet. The ratio of escape speed from planet to escape speed from moon is ________.

alok kumar singh

-> $V_{e s c} = \sqrt[]{\frac{2 G M}{R}}$

$R a t i o = \sqrt[]{\frac{81}{9}} = 3$

If the gravitational acceleration at the surface of earth is g, then increase in potential energy in lifting an object of mass m from earth surface to a height equal to half of radius of earth from surface will be:

alok kumar singh