Weightlessness - class-XI

Mass of an object always remains constant whether the object is accelerating or not. Apparent weight of the object changes due to acceleration.

In the train frame, pseudo force ma acts down the plane. Resolving forces perpendicular and parallel to the ceiling gives the effective angle of the string with the normal to the ceiling. With acceleration a = g/2 up an incline of 30 degrees, the angle evaluates correctly to tan^(-1)(2 / sqrt(3)).

When the lift moves $UP$ with uniform acceleration $a$ $Pseudo$ force on the man is $ma$ downward.

When the train is either at rest or moving with constant velocity, then the acceleration of the train is zero and hence there is no pseudo force acting on the stone in the horizontal direction which makes the string tied to the stone to stay in vertical direction.

This happens due to inertia of rest . initially both car and people are in rest , when car accelerates rapidly , feet of the people also accelerates with car because they are in contact with the floor of car , but heads of the people do not accelerates quickly because they want to be in rest due inertia of rest therefore people feel a jerk .

Since cage is at rest no pseudo force will be acting on bird hence weight of bird will be $2kg$ in air, therefore weight of bird and cage assembly will be  equal to 2kg and 1kg i.e equal to $3kg$.

Reading shown on the spring balance is weight of bird + weight of cage. When bird starts flying normal force exerted by bird on the case become zero. When bird flies his weight is carried by air but case in made of wire so no additional force of weight of bird act on the cage so reading on the cage is less when bird is flying.

Pseudo force is defined as F_p = -m * a_frame. If a_frame = 0 (inertial frame), then F_p = 0. If the frame is non-inertial (a != 0), the pseudo force is non-zero. It is impossible for a frame to be both inertial and non-inertial simultaneously, so F1=0 and F2=0 is impossible if S2 is accelerating relative to S1.

As the apple is dropped, it is under free-fall meaning that the force of gravity is acting on it. With respect to the person inside the lift, the apple seems not to be falling Hence, the man and the lift must also be falling with the action of acceleration due to gravity i.e, a constant acceleration.

When lift accelerates up, pseudo force acts downwards, hence it increases apparent weight. $W _{1}>mg$
When lift is stationary, $W _{2}=mg$
When lift falls freely, it accelerates with g downwards, causing an upwards pseudo force in the frame of the lift equal to mg. Hence total force is 0. So weight is 0. $W _{3}=0$
When lift accelerates down, pseudo force acts upwards,  hence it decreases apparent weight $W _{4}<mg$, but also the acceleration is less than g, therefore $W _{4}=m(g-a)>0$

Hence, $W _{3}<W _{4}<W _{2}<W _{1}$

Working in the frame of universe:

The mass of the lift is $1000\;{\rm{kg}}$, the acceleration is $1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ and the acceleration due to gravity is $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

The tension developed in string is given as,

$T = mg + ma$

$ = 1000 \times 9.8 + 1000 \times 1$

$ = 10800\;{\rm{N}}$

Thus, the tension developed in the string is $10800\;{\rm{N}}$.

An astronaut experiences weightlessness in a space satellite. It is because the astronaut experiences no gravity.

As the astronauts are in a frame which is rotating around the earth and the centripetal acceleration on astronauts and the satellite is $same$, $g=GM/r^2$ where $r$ is the radius of the orbit. So relative to the satellite the astronauts can't exert any force on satellite so no reaction force from the satellites so they feel the condition of weightlessness.

Weightlessness occurs when the normal force is zero. At the center of the Earth, gravitational force is zero, but you are not in free fall in the same sense as an orbit; however, you would experience zero weight. Wait, actually, in a lift moving with uniform velocity, the acceleration is zero, so the normal force equals your weight (mg), meaning you feel your normal weight. Therefore, you do NOT experience weightlessness in a lift moving with uniform velocity.

An astronaut,inside an earth`s satellite experiences weightlessness because he is falling freely.

Both statements are true. An artificial satellite is in free fall, leading to weightlessness. On the Moon, you are standing on a surface, so the normal force balances the gravitational pull, meaning you feel weight. Statement 2 is a true statement, but it doesn't explain the reason for the difference in the sensation of weightlessness (which is about the presence of a normal force, not just the source of gravity).

Weight of a body is due to its weight. A body becomes weightless when effective gravity on the body is zero.

In a spaceship the astronaut weighs zero. Since there is no gravity in the space person feels weightless. Apart from these space is a vacuum and there is no air resistance and thus we feel weightless.

The satellite rotates in a circular path or elliptical path such that gravitational pull is always balanced by centrifugal force due to the rotational motion of satellite.

The satellite moves in the orbit with a constant speed in circular path, with centripetal acceleration being provided by the force of gravity on it.

A geostationary satellite revolves around the earths with the same time period of earth's rotation, that is 24 hours.
Since it revolves with same speed, the relative velocity is zero with respect to earth and hence any body inside a geostationary satellite doesn't feel the gravity.
So, the weight would be equal to zero.

Inside a satellite, every object experiences weightlessness

Time period of simple pendulum is given by:
$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$
where l is the length of the pendulum.
Inside a satellite, $g = 0$
Hence, period will be infinite which means there will be no oscillation.

Weightlessness at the equator occurs when the centripetal acceleration equals gravity: omega^2 * R = g. Current omega = sqrt(g/R). To feel weightless, omega' = sqrt(g/R). Wait, the current angular velocity is such that omega^2 * R is much smaller than g. The required increase is to make omega'^2 * R = g. The ratio omega'/omega = sqrt(g/R) / (v/R) = ... The percentage increase is approximately 17%.

In an elliptical orbit, the gravitational force is central, so angular momentum and areal velocity (Kepler's second law) are conserved. Total energy is also conserved. However, the velocity vector changes in both magnitude and direction, so linear momentum is not constant.

As a person sits in a chair $($ on Earth's surface $)$, he will experience two forces, the force of the Earth's gravitational field pulling him downward toward the Earth and the force of the chair pushing him upward. The upward chair force is sometimes referred to as a normal force.
If there were no upward normal force acting upon body, body would not have any sensation of the weight. Without the contact force $($ the normal force $)$, there is no means of feeling the non-contact force $($ the force of gravity $)$.
Hence, A person sitting in a chair in a satellite feels weightless because normal force is zero.

On a rotating body, as we move away from center of rotation the centrifugal force increases. So, the centrifugal force is maximum at the periphery of a rotating wheel. Thus, having the astronaut room there would solve some of the weightlessness problem in satellite.

time period $\propto \sqrt { \cfrac { l }{ g }  } $ , as in the satellite there will be no gravity so the time period will be infinite.(logical explanation. : without gravity there will be no force on pendulum so it will not move a bit even in infinite time.)

In the satellite the spring is in free fall so no weight will be recorded.

Orbital speed v = sqrt(GM/R). Mass M = density * volume = p * (4/3) * pi * R^3. Substituting M: v = sqrt(G * p * 4/3 * pi * R^3 / R) = sqrt(4/3 * pi * G * p * R^2) = 2R * sqrt(pi * G * p / 3).

For an earth satellite moving in a circular orbit, centripetal force required for its circular motion is provided by the gravitational force exerted by earth on it.
 It means resultant force on the astronaut is equal to the gravitational force exerted by earth on him. Hence, no reaction is exerted by floor of the satellite on him.
In other words, his acceleration towards earth centre (centripetal acceleration) is exactly equal to acceleration caused by the gravitational force alone.
Hence, options (b) and (c) are correct.

Weightlessness means that there is no reaction on a body from the floor. Since both the artificial satellite & the astronaut have same centripetal acceleration (as in a lift; which is falling freely, we does not feel any weight, because both lift & we fall with same acceleration). so the astronaut does not feel any weight inside the space craft

$\displaystyle T=2\pi \sqrt {\frac{l}{g}}$
but inside the satellite $\displaystyle g=0$
So $\displaystyle T=\infty$ 

Since,Both the satellites are freely falling bodies.
so,$W _1=W _2$

As Moon has its mass, thus it posses its own gravity ($\frac {1}{6}$th of that of the Earth). Hence, weightlessness is not experienced at the surface of the Moon.

In a satellite, weightlessness is felt.

Answer is A.

The reading on the spring balance is independent of the radius and thus both weight will be the same. Also, there is no gravitational force acting on a satellite.
 
Hence, ${ W } _{ 1 }={ W } _{ 2 }$.

The gravitational force acting on the man in the satellite will be zero as there is no gravitational pull on man in the space hence his weight will be zero.

Satellite going around the earth in circular orbit is in state of free fall, and its speed is constant. speed depends upon the radius of orbit of satellite.

Simulated earth gravity can be realized by rotating the space station about a central axis. This rotation creates centrifugal force on the people inside the space station away from the central axis. Thus, $g={ \omega  }^{ 2 }R $ at the central axis $R=0$. So, gravitational field is zero.

For a man at an angle $\theta=60^o$

$0=g-{ Rw }^{ 2 }\cos ^{ 2 }{ 60° } \ { w }^{ 2 }=\cfrac { 4g }{ R } \quad or,{ w }^{ 2 }=2\sqrt { \cfrac { g }{ R }  } \ \cfrac { 2\pi  }{ T } =2\sqrt { \cfrac { g }{ R }  } \ \therefore T=\pi \sqrt { \cfrac { R }{ g }  } $

For the body to follow circular path, there must be a centripetal force. Here the astronaut is inside a satellite which is revolving around the earth under the influence of earth's gravitation. Thus, the earth's gravitation acts as centripetal force and the net force on astronaut is zero.
Statement 2 is right explanation of 1.