JEE Challenger
More from Gravitation

Astronaut Weightlessness in Orbiting Space Station and Free Fall

STATEMENT-1
An astronaut in an orbiting space station above the Earth experiences weightlessness.
and
STATEMENT-2
An object moving around the Earth under the influence of Earth's gravitational force is in a state of 'free-fall'.

Options

A

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

Correct
B

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

C

STATEMENT-1 is True, STATEMENT-2 is False

D

STATEMENT-1 is False, STATEMENT-2 is True

Topics & Concepts

Step-by-Step Solution

To determine the correctness of the statements and their relationship, we analyze the physics of an orbiting body:

  1. Analysis of STATEMENT-2:
    By definition, an object is said to be in a state of free-fall if the only force acting upon it is gravity.
    An orbiting satellite or space station moves in a circular or elliptical path around the Earth solely under the influence of the Earth's gravitational force.
    The gravitational force provides the necessary centripetal force:
    Fg=GMEmr2=macF_g = \frac{G M_E m}{r^2} = m a_c
    Therefore, the acceleration of the object is strictly equal to the local acceleration due to gravity:
    a=g(r)=GMEr2a = g(r) = \frac{G M_E}{r^2}
    Since no other non-gravitational force (like thrust or aerodynamic drag) acts on the object, it is continuously falling towards the Earth while moving forward, which means it is in a perpetual state of free-fall.
    Thus, STATEMENT-2 is True.

  2. Analysis of STATEMENT-1:
    Consider an astronaut of mass mm standing on the floor of the orbiting space station. Both the astronaut and the space station are subject to the same gravitational field at distance rr from the Earth's center:
    astation=g(r)andaastronaut=g(r)a_{\text{station}} = g(r) \quad \text{and} \quad a_{\text{astronaut}} = g(r)
    Applying Newton's second law to the astronaut in the radial direction:
    mg(r)−N=maastronautm g(r) - N = m a_{\text{astronaut}}
    Since the astronaut accelerates along with the space station at aastronaut=g(r)a_{\text{astronaut}} = g(r), we have:
    mg(r)−N=mg(r)  ⟹  N=0m g(r) - N = m g(r) \implies N = 0
    Here, NN is the normal contact reaction exerted by the floor of the space station on the astronaut.
    Apparent weight is defined as the magnitude of the contact/support force acting on a body. Since N=0N = 0, the astronaut experiences an apparent weight of zero, which is the state of weightlessness.
    Thus, STATEMENT-1 is True.

  3. Relationship between the statements:
    The reason the normal force NN is zero (causing the sensation of weightlessness) is precisely because both the astronaut and the station are falling together purely under the influence of gravity—that is, they are both in a state of free-fall.
    Therefore, STATEMENT-2 is the correct scientific explanation for STATEMENT-1.

Hence, the correct option is (A).