An astronaut goes out for a “space-walk” at a distance above the earth equal to the radius of the earth. What is her acceleration due to gravity at that point? zero g g/2 g/4
The correct answer and explanation is:
The correct answer is: g/4
Explanation:
The acceleration due to gravity g at the surface of the Earth is given by the formula: g=GMR2g = \frac{G M}{R^2}
Where:
- GG is the universal gravitational constant
- MM is the mass of the Earth
- RR is the radius of the Earth
Now, when the astronaut is at a distance above the Earth equal to the Earth’s radius, her distance from the center of the Earth becomes: r=R+R=2Rr = R + R = 2R
So, her new acceleration due to gravity g′g’ becomes: g′=GM(2R)2=GM4R2=g4g’ = \frac{G M}{(2R)^2} = \frac{G M}{4R^2} = \frac{g}{4}
Conceptual Understanding:
Gravity decreases with the square of the distance from the center of the Earth. If you move twice as far from the center, the gravitational pull becomes one-fourth as strong.
- At 1 Earth radius above the surface (i.e., total distance = 2R), gravity isn’t zero — it just gets weaker.
- The idea of “zero gravity” in space is a misconception. Astronauts in orbit still experience gravity — it’s just weaker, and they’re in free fall, creating the sensation of weightlessness.
Conclusion:
At a distance equal to the Earth’s radius above the surface, the astronaut is at 2R from Earth’s center. Hence, her acceleration due to gravity becomes g/4, not zero.
This reduction follows the inverse square law of gravitation.