Problem Statement
Solve the gravitation problem: Solve the Newton’s Laws / mechanics problem: Mass $m = 1$ kg is placed at the origin. Two masses $M = 10$ kg each are placed at $(0.5, 0)$ m and $(-0.5, 0)$ m. Find the net gravitational force on $m$. ($G = 6.67\times10^{-11}$ N m² kg⁻²) Gravitational force is attractive; forces from equal masses at
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
$$\frac{mv^2}{r} = \frac{GMm}{r^2} \implies v = \sqrt{\frac{GM}{r}}$$
$$T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi\sqrt{\frac{r^3}{GM}}$$
$$\frac{1}{2}mv_e^2 = \frac{GMm}{R} \implies v_e = \sqrt{\frac{2GM}{R}}$$
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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