Numerical 1 — Complete Satellite Analysis at h = R

Given:

• Height: h = R = $6.4 \times 10^{6}$ m
• Satellite mass: m = 200 kg
• g = 10 m $s^{-2}$
• R = $6.4 \times 10^{6}$ m

Step 1: Find orbital radius r

$r = R + h = R + R = 2R = 2 \times 6.4 \times 10^6\ \text{m} = 1.28 \times 10^7\ \text{m}$

Step 2: Find orbital velocity v_{0}

$v_0 = \sqrt{\frac{gR^2}{r}} = \sqrt{\frac{gR^2}{2R}} = \sqrt{\frac{gR}{2}} = \sqrt{\frac{10 \times 6.4 \times 10^6}{2}}$

$= \sqrt{3.2 \times 10^7}\ \text{m s}^{-1} = 5{,}657\ \text{m s}^{-1} \approx 5.66\ \text{km s}^{-1}$

Step 3: Find orbital period T

$T = \frac{2\pi r}{v_0} = \frac{2\pi \times 1.28 \times 10^7\ \text{m}}{5{,}657\ \text{m s}^{-1}} = \frac{8.04 \times 10^7}{5{,}657}\ \text{s}$

$= 14{,}210\ \text{s} = \frac{14{,}210}{3600}\ \text{h} \approx 3.95\ \text{h}$

Step 4: Find KE, PE, Total Energy

$KE = \frac{1}{2}mv_0^2 = \frac{1}{2} \times 200 \times (5657)^2 = 100 \times 3.2 \times 10^7 = 3.2 \times 10^9\ \text{J} = 3.2\ \text{GJ}$

$PE = -\frac{gR^2 m}{r} = -\frac{gR^2 m}{2R} = -\frac{gRm}{2} \times 2 = -\frac{10 \times 6.4 \times 10^6 \times 200}{1} = ...$

Wait — correct formula: $PE = -GMm/r = -gR^2 m/r = -gR^2m/(2R) = -gRm/2$

$PE = -\frac{10 \times 6.4 \times 10^6 \times 200}{2} \times \frac{1}{1} = -10 \times 6.4 \times 10^6 \times 100$

$PE = -6.4 \times 10^9\ \text{J} = -6.4\ \text{GJ}$

Verify: |PE| = 2 × KE = 2 × 3.2 = 6.4 GJ ✓

$E_{\text{total}} = KE + PE = 3.2 - 6.4 = -3.2\ \text{GJ}$

Verify: E = −KE = −3.2 GJ ✓

Summary: