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.

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 $\boxed{7900 \text{ m/s}}$.

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 $\boxed{85 \text{ min}}$.