Why does an astronaut in an orbiting space station experience weightlessness even though $g$ is non-zero at that altitude?

Weightlessness is not the absence of gravity, but the absence of a contact normal reaction force. Both the space station and the astronaut accelerate toward Earth at the exact same field acceleration $g' = \frac{G M}{r^2}$. Because there is zero relative acceleration between the astronaut and the cabin floor, the normal reaction force $N = 0$, producing the sensation of weightlessness.

How do I choose between Angular Momentum Conservation and Energy Conservation in satellite mechanics?

Use both simultaneously for elliptical orbits. Conservation of Angular Momentum $L = m v_1 r_1 \sin\theta_1 = m v_2 r_2 \sin\theta_2$ relates speeds and distances specifically at periapsis and apoapsis where velocity is perpendicular to radius vector ($\ me\sin 90^\circ = 1$). Use Conservation of Mechanical Energy $E = \frac{1}{2}m v^2 - \frac{GMm}{r}$ alongside it to solve for the two unknown state variables ($v$ and $r$).

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Question #1NEET 2024

A body of mass $m$ is projected vertically upwards from the surface of Earth ( $Radius = R$ ) with an initial velocity $v = \frac{1}{2} v_e$ , where $v_e$ is the escape velocity from Earth. The maximum height $h$ attained by the body above the surface of Earth is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Write total initial energy at Earth's surface. Since $v = \frac{1}{2} v_e = \frac{1}{2}\sqrt{\frac{2GM}{R}}$ , initial kinetic energy is $K_i = \frac{1}{2}m v^2 = \frac{1}{2}m \left(\frac{GM}{2R}\right) = \frac{GMm}{4R}$ . Initial total energy is $E_i = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}$ .

Step 2

At maximum height $h$ , velocity drops to zero, so $K_f = 0$ . Total mechanical energy at height $h$ is $E_f = -\frac{GMm}{R+h}$ .

Step 3

Equate initial and final mechanical energy: $-\frac{3GMm}{4R} = -\frac{GMm}{R+h} \implies R+h = \frac{4}{3}R \implies h = \frac{R}{3}$ .

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Question #2JEE Main 2023

Two spherical stars of masses $M$ and $16M$ have radii $R$ and $2R$ respectively. Their centers are separated by a distance $10R$ . A body of mass $m$ is projected from the surface of the larger star directly towards the smaller star. The minimum initial speed $v_{min}$ required for the mass to reach the surface of the smaller star is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Locate the neutral point $N$ where net gravitational field is zero between the centers. Let distance of $N$ from center of mass $M$ be $x$ . Then $\frac{G M}{x^2} = \frac{G (16 M)}{(10R - x)^2} \implies \frac{1}{x} = \frac{4}{10R - x} \implies x = 2R$ . Thus, neutral point $N$ is at distance $2R$ from mass $M$ and $8R$ from mass $16M$ .

Step 2

Apply Energy Conservation between surface of larger star (center distance $2R$ from $16M$ , $8R$ from $M$ ) and neutral point $N$ (center distance $8R$ from $16M$ , $2R$ from $M$ ). Initial potential energy $U_i = -\frac{G(16M)m}{2R} - \frac{GMm}{8R} = -\frac{65 GMm}{8R}$ . Potential energy at $N$ is $U_N = -\frac{G(16M)m}{8R} - \frac{GMm}{2R} = -\frac{20 GMm}{8R}$ .

Step 3

To cross $N$ , minimum kinetic energy at start must satisfy $\frac{1}{2}m v_{min}^2 + U_i = U_N \implies \frac{1}{2}m v_{min}^2 = U_N - U_i = \frac{45 GMm}{8R} \implies v_{min} = \sqrt{\frac{45 GM}{4R}}$ .

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Mastering Gravitation: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Universal Law of Gravitation

Variation of $g$ with Height and Depth

Gravitational Potential Energy

Escape Velocity

Orbital Velocity & Time Period

Effective Acceleration Due to Earth's Rotation

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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