To estimate the speed and period of a satellite close to the Earth's surface, we use the given radius of orbit and standard physical constants. Given: • Radius of orbit, $r = 6.4 \times 10^6 \text{ m}$ (assumed to be equal to Earth's radius) • Acceleration due to gravity at Earth's surface, $g = 9.8 \text{ m/s}^2$ (a) Estimate the speed of the satellite. Step 1: Use the formula for orbital speed for a satellite orbiting close to the Earth's surface. The orbital speed $v$ is given by: $$ v = \sqrt{gr} $$ Substitute the given values: $$ v = \sqrt{(9.8 \text{ m/s}^2) \times (6.4 \times 10^6 \text{ m})} $$ $$ v = \sqrt{62.72 \times 10^6 \text{ m}^2/\text{s}^2} $$ $$ v = \sqrt{6.272 \times 10^7 \text{ m}^2/\text{s}^2} $$ $$ v \approx 7919.6 \text{ m/s} $$ Rounding to two significant figures: $$ v \approx 7900 \text{ m/s} $$ The speed of the satellite is $\boxed{7900 \text{ m/s}}$. (b) Estimate the period of the satellite. Step 2: Use the formula for the period of orbit. The period $T$ is the time taken for one complete orbit, given by: $$ T = \frac{2\pi r}{v} $$ Substitute the values for $r$ and the calculated $v$: $$ T = \frac{2\pi (6.4 \times 10^6 \text{ m})}{7919.6 \text{ m/s}} $$ $$ T = \frac{4.0212 \times 10^7 \text{ m}}{7919.6 \text{ m/s}} $$ $$ T \approx 5077.7 \text{ s} $$ Rounding to two significant figures: $$ T \approx 5100 \text{ s} $$ To convert to minutes: $$ T = 5077.7 \text{ s} \times \frac{1 \text{ min}}{60 \text{ s}} $$ $$ T \approx 84.6 \text{ min} $$ Rounding to two significant figures: $$ T \approx 85 \text{ min} $$ The period of the satellite is $\boxed{5100 \text{ s}}$ or $\boxed{85 \text{ min}}$.