1. Given that the acceleration due to gravity decays with height from the centre of the Earth following an inverse square law, what is the percentage change in g from the Earth's surface to an altitude of 100 km? (See also Problem 6 of Chapter 3.)

FIGURE 1.8. A photograph of the sound barrier being broken by a US Navy jet as it crosses the Pacific Ocean at the speed of sound just 75 feet above the ocean. Condensation of water is caused by the rapid expansion and subsequent adiabatic cooling of air parcels induced by the shock (expansion/compression) waves caused by the plane outrunning the sound waves in front of it. Photograph was taken by John Gay from the top of an aircraft carrier. The photo won First Prize in the science and technology division of the World Press Photo 2000 contest.

FIGURE 1.8. A photograph of the sound barrier being broken by a US Navy jet as it crosses the Pacific Ocean at the speed of sound just 75 feet above the ocean. Condensation of water is caused by the rapid expansion and subsequent adiabatic cooling of air parcels induced by the shock (expansion/compression) waves caused by the plane outrunning the sound waves in front of it. Photograph was taken by John Gay from the top of an aircraft carrier. The photo won First Prize in the science and technology division of the World Press Photo 2000 contest.

2. Compute the mean pressure at the Earth's surface given the total mass of the atmosphere, Ma (Table 1.3), the acceleration due to gravity, g, and the radius of the Earth, a (Table 1.1).

3. Express your answer to Problem 2 in terms of the number of apples per square meter required to exert the same pressure. You may assume that a typical apple weighs 0.2 kg. If the average density of air is 5 apples per m3 (in apple units), calculate how high the apples would have to be stacked at this density to exert a surface pressure equal to 1000 hPa. Compare your estimate to the scale height, H, given by Eq. 3.6 in Section 3.3.

4. Using (i) Eq. 1-4, which relates the saturation vapor pressure of H2O to temperature T, and (ii) the equation of state of water vapor, e = pvRvT (see discussion in Section 1.3.2), compute the maximum amount of water vapor per unit volume that air can hold at the surface, where Ts = 288 K, and at a height of 10 km where (from Fig. 3.1) Tio km = 220 K. Express your answer in kgm-3. What are the implications of your results for the distribution of water vapor in the atmosphere?

Was this article helpful?

This is a product all about solar power. Within this product you will get 24 videos, 5 guides, reviews and much more. This product is great for affiliate marketers who is trying to market products all about alternative energy.

## Post a comment