Gravitation — Flashcards
Telangana Open School (TOSS) · Class 12 · Physics
30 flashcards for Gravitation (Telangana Open School (TOSS) Class 12 Physics) to test yourself on key terms and facts.
Interactive on Super Tutor
Studying Gravitation? Get the full interactive chapter.
Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for flashcards and more.
Free trial, no card needed.
State Newton's Law of Universal Gravitation.
Answer
Every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Formula: \…
Calculate the gravitational force between two masses of 100 kg and 200 kg separated by 2 m.
Answer
Given: \( m_1 = 100 \, \text{kg}, m_2 = 200 \, \text{kg}, r = 2 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \) Step 1: Use formula \( F = G \frac{m_1 m_2}{r^2} \) Step 2: \( F…
Why does an apple fall to the ground but the Moon does not?
Answer
Both are pulled by Earth's gravity. The apple falls straight because it has no sideways speed. The Moon has high tangential velocity, so it keeps 'falling' around Earth in orbit. This is like throwing…
Derive the expression for acceleration due to gravity (g) on Earth's surface.
Answer
Gravitational force on mass m: \( F = G \frac{M m}{R^2} \) By Newton's second law: \( F = m g \) So, \( m g = G \frac{M m}{R^2} \) Cancel m: \( g = G \frac{M}{R^2} \) Where M = Earth's mass, R = E…
Calculate the value of 'g' on Earth using M = 5.97 × 10²⁴ kg and R = 6.37 × 10⁶ m.
Answer
Given: \( M = 5.97 \times 10^{24} \, \text{kg}, R = 6.37 \times 10^6 \, \text{m}, G = 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \) Formula: \( g = G \frac{M}{R^2} \) Step 1: \( R^2 = (6.37 \…
How does 'g' vary with height above Earth's surface?
Answer
As height (h) increases, 'g' decreases. For small h compared to R, \( g_h = g \left(1 - \frac{2h}{R}\right) \). This is because gravity weakens with square of distance from Earth's center. Example: At…
Find the value of 'g' at a height of 1000 km above Earth's surface.
Answer
Given: \( h = 1000 \, \text{km} = 10^6 \, \text{m}, R = 6.4 \times 10^6 \, \text{m}, g = 9.8 \, \text{ms}^{-2} \) Formula: \( g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \) Step 1: \( \frac{h}{R}…
Why does 'g' decrease with depth inside Earth?
Answer
As we go deeper, only the mass below that point contributes to gravity. The outer shell exerts no net gravitational force. So, effective mass attracting decreases, reducing 'g'. At Earth's center, g =…
+22 more flashcards
Practise AllFrequently Asked Questions
What are the important topics in Gravitation for Telangana Open School (TOSS) Class 12 Physics?
How many flashcards are available for Gravitation?
How should I revise Gravitation for the Telangana Open School (TOSS) Class 12 board exam?
Sources & Official References
Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.
More resources for Gravitation
Practice Quiz
Test yourself with a quick quiz
Important Questions
Exam-style questions with answers
Revision Notes
Key points for last-minute revision
Formula Sheet
The chapter's formulas in one place
Chapter Summary
Understand the chapter at a glance
Concept Maps
See how topics connect
Study Plan
Step-by-step plan for this chapter
Syllabus
What topics to cover
For serious students
Get the full Gravitation chapter — start free.
Quizzes, flashcards, an AI doubt solver and a study plan for Telangana Open School (TOSS) Class 12 Physics. Free to start, no card needed.