Physics Topics can help us understand the behavior of the natural world around us.

## What is Apparent Weight?

It is known that the weight of a body of mass m is mg, i.e., the earth pulls the body towards its centre with this force. When this body is placed on a non-accelerating plane, it exerts a force of mg on that plane.

The plane also exerts an upward reaction force on the body R(= mg). This upward reaction force is the apparent weight of the body. The body feels only this upward reaction force as its weight. Hence, any person or body cannot feel its own weight while standing on a plane if the plane does not offer any upward reaction; this means that the person or body feels weightless. Let the plane along with the body have an acceleration vertically upwards. In this case, the upward reaction of the plane does not become equal to the weight.

Because of this a passenger in a lift with an acceleration feels his weight to be increased or decreased (see the chapter Newton’s Laws of Motion). There are a few cases in which the apparent weight of a body is not related to the existence of the plane (on which the body is placed). But it is convenient to measure the apparent weight of a body taking into consideration the existence of such a plane.

Weight experienced by on astronaut before the spaceship is parked in its orbit: The velocity of the spaceship is increased rapidly by the rocket immediately after its launch from the surface of the earth [Fig.(a)], Sometimes the value of its upward acceleration may be 15 times the value of the acceleration due to gravity. If the mass of the astronaut is m, the upward reaction of the spaceship on the man is R, and the acceleration of the spaceship is 15g, then

R – mg = ma = m ᐧ 15g or, R = 16 mg

This is the apparent weight of the astronaut. Thus he feels himself to be 16 times heavier than his actual weight.

Weight experienced by an astronaut when the spaceship is moving in its orbit:

Let r = radius of the circular orbit of the spaceship with respect to the centre of the earth [Fig.(b)],

g’ = acceleration due to gravity at the position of the spaceship,

v = instantaneous velocity of the spaceship,

M = mass of the spaceship, m = mass of the astronaut,

R = upward reaction of the floor of the spaceship on the astronaut

For the rotation of the spaceship or of the astronaut around the earth, the force acting towards the centre of the earth supplies the necessary centripetal force. So, for the spaceship and for the astronaut, respectively,

Mg’ = \(\frac{M v^2}{r}\) or g’ = \(\frac{v^2}{r}\)

and mg’ – R = \(\frac{m v^2}{r}\) or, m\(\frac{v^2}{r}\) – R = \(\frac{m v^2}{r}\)

Thus, R = 0

The astronaut cannot feel his own weight as the reaction force is zero. He floats in the spaceship. This is called the weightlessness of man or any object in an artificial satellite.

Similarly, if it is assumed that the spaceship is resting on an imaginary plane and the motion of the spaceship along with its plane is considered, the spaceship also appears to be weightless [Fig.(c)], As the plane is imaginary, it can be concluded that all objects (planet, satellite, etc.) orbiting under the action of gravitation do not experience any force directed towards the centre of the orbit, and therefore, all such objects are weightless.

An astronaut feels much heavier than his actual weight at the time of launching the spaceship, but he feels weightless as soon as the spaceship is placed in its orbit.

The phenomenon of weightlessness is widely used in many cases of scientific and technological requirements. For example, crystal formation on the earth’s surface due to gravitation is imperfect, but perfect crystals can be developed in orbiting spaceships easily and such crystals have immense importance in making transistors and other such electronic devices. Again, oil and water cannot really be mixed on the earth’s surface, but in a spaceship, in the weightless condition, it is possible to make a homogeneous mixture of oil and water. In addition, a pure alloy can be produced by cooling a homogeneous mixture of molten metals in a spaceship.

As in the case of materials kept in an artificial satellite, objects on the moon’s surface also have no weight due to the gravitational pull of the earth. But no body is weightless on the moon’s surface as the attraction of the moon acts on the body. This weight is about \(\frac{1}{6}\) th of its weight on the earth. It is to be noted that bodies in an artificial satellite are also subject to the force of attraction of the satellite or spaceship. But this force is too small to be felt.

The gravitational force of attraction of the earth on an artificial satellite or on an astronaut in it can never be zero. If it had been zero, the necessary centripetal force could not have been supplied and the satellite would not have been able to revolve around the earth in a circular orbit. Thus the statement that a body is weightless implies that only the reaction force acting on it is zero, but the gravitational force is not zero.

### Numerical Examples

**Example 1.**

Two particles of equal mass revolve in a circular path of radius R due to their mutual force of attraction. Find the velocity of each particle.

**Solution:**

At any time, during the revolution, the two particles stay at the two ends of any diameter of the circular path. Mutual force of gravitation acts along the diameter. Considering the motion of any one of the particles,

\(\frac{G m \times m}{(2 R)^2}\) = \(\frac{m v^2}{R}\) or, v^{2} = \(\frac{G m}{4 R}\) or, v = \(\sqrt{\frac{G m}{4 R}}\).

**Example 2.**

The acceleration due to gravity at two places are g and g’ respectively. A body is dropped from the same height at both places. At the second place, the required time to touch the ground is t seconds less than that in the first place, while the velocity attained in reaching the ground is higher by a value of v than that in the first place. Show that gg’ = \(\frac{v^2}{t^2}\).

**Solution:**

Let the time taken by the body to fall at the first place be T and velocity with which the body touches the ground be V.

Hence, the time taken to reach the ground and the corre-sponding velocity are (T – t) and V + v in the second place. Let the body fall from a height h in each case. Considering the motion at the first place,

**Example 3.**

A satellite is orbiting in a circular path around the earth close to its surface. What additional velocity is to be imparted to the satellite so that it escapes the earth’s gravitational pull? The radius of the earth = 6400 km and g = 9.8 m ᐧ s^{-2}.

**Solution:**

Angular momentum of the satellite about the centre of the orbit is L = mvr, where v is the orbital speed of the satellite. For motion in a fixed orbit,

centripetal force = force of attraction due to gravity

or, \(\frac{m v^2}{r}\) = \(\frac{G M m}{r^2}\)

or, mv^{2}r = GMm or, (mvr)^{2} = GMm^{2}r

or, mvr = m\(\sqrt{G M r}\)

∴ L = m\(\sqrt{G M r}\)

**Example 4.**

A satellite is orbiting in a circular path around the earth close to its surface. What additional velocity is to be imparted to the satellite so that it escapes the earth’s gravitational pull? The radius of the earth = 6400 km and g = 9.8 m ᐧ s^{-2}.

**Solution:**

Orbital speed of a satellite in an orbit close to the surface of the earth v = \(\sqrt{g R}\); escape velocity v_{e} = \(\sqrt{2 g R}\)

∴ Required additional velocity

= v_{e} – v = \(\sqrt{2 g R}\) – \(\sqrt{g R}\)

= (\(\sqrt{2}\) – 1 )\(\sqrt{g R}\) = 0.41 × \(\sqrt{9.8 \times 6400 \times 1000}\)

= 3.25 × 10^{3} m ᐧ s^{-1} = 3.25 km ᐧ s^{-1}.

**Example 5.**

Two satellites A and B of the same mass revolve around the earth in circular orbits. They are at heights of R and 3R respectively from the surface of the earth (R = radius of the earth). Find the ratio of their kinetic energies and potential energies.

**Solution:**

Height of satellite A from the centre of the earth = R+ R = 2R and height of satellite B from the centre of the earth = R + 3R = 4R. If the mass of each satellite is m and the mass of the earth is M,

potential energy of satellite A, U_{A} = –\(\frac{G M m}{2 R}\)

and potential energy of satellite B, U_{B} = –\(\frac{G M m}{4 R}\)

∴ \(\frac{U_A}{U_B}\) = \(\frac{G M m / 2 R}{G M m / 4 R}\) = \(\frac{2}{1}\)

If orbital speeds of the two satellites are v_{1} and v_{2}, then kinetic energy of satellite A, K_{A} = \(\frac{1}{2} m v_1^2\);

and kinetic energy of satellite B, K_{B} = \(\frac{1}{2} m v_2^2\)

As v_{1} = \(\sqrt{\frac{G M}{2 R}}\) and v_{2} = \(\sqrt{\frac{G M}{4 R}}\)

∴ \(\frac{K_A}{K_B}\) = \(\frac{v_1^2}{v_2^2}\) = \(\frac{G M / 2 R}{G M / 4 R}\) = \(\frac{2}{1}\).

**Example 6.**

Two satellites S_{1} and S_{2} are orbiting around the earth in circular orbits in the same direction. Time period for the two satellites are 1h and 8h respectively. The radius of the orbit of satellite S_{1} is km. If satellites S_{1} and S_{3} are on the same side of the earth, find the linear and angular speeds of S_{2 }with respect to S_{1}.

**Solution:**

If the radii of satellites S_{1} and S_{2} are r_{1} and r_{2} and their respective time periods are T_{1} and T_{2}, then

∴ Linear speed of S_{2} with respect to S_{1}

= |v_{2} – v_{1}| = π × 10^{4}km ᐧ h^{-1}

and angular speed of S_{2} with respect to S_{1},

ω = \(\frac{\left|v_2-v_1\right|}{r_2-r_1}\) = \(\frac{\pi \times 10^4}{4 \times 10^4-10^4}\)

= \(\frac{\pi}{3}\)radᐧs^{-1}