Problem Statement
Solve the gravitation problem: Solve the Newton’s Laws / mechanics problem: Find the gravitational force between two masses of 100 kg and 200 kg separated by 1 m. ($G = 6.67 \times 10^{-11}$ N m² kg⁻²) $F = Gm_1m_2/r^2$ Step 1: $F = \dfrac{6.67\times10^{-11}\times100\times200}{1^2} = 6.67\times10^{-11}\times20000 = 1.334\times10^
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Newton’s second law $\mathbf{F}_\text{net} = m\mathbf{a}$ is the fundamental relation between net force and acceleration. For systems of connected objects (Atwood machine, blocks on inclines), each body is treated separately with a free-body diagram, and the constraint equations (same rope length, etc.) link the accelerations.
- $\mathbf{F}_{\text{net}} = m\mathbf{a}$ — Newton’s second law
- Atwood: $a = (m_1-m_2)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$
- $f_k = \mu_k N$ — kinetic friction
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
Full substitution shown in the steps above.
Answer
$$\boxed{v_e = \sqrt{2GM/R} \approx 11.2\,\text{km/s}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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