By defining after-tax income, demonstrate the difference resulting from a $500 tax credit versus a $500 tax deduction for a single taxpayer in the 25% tax bracket with $41,000 of pre-tax income.

Using the formula for gravitational force {eq}F_g = G \frac{m_1 m_2}{r^2} {/eq}, where {eq}G{/eq} is the gravitational constant, {eq}m_1{/eq} and {eq}m_2{/eq} are the masses of the two objects, and {eq}r{/eq} is the distance between them. Assuming that the mass of the Earth is much larger compared to the mass of the person, we can assume that the person's weight is due to the gravitational force of the Earth. At sea level, the distance from the person to the center of the Earth is {eq}r_1 = 6.38 times 10^6 m + h_1 {/eq}, where {eq}h_1 = 0 {/eq} is the height of sea level. At an altitude of {eq}h_2 = 5000 m{/eq}, the distance from the person to the center of the Earth is {eq}r_2 = 6.38 times 10^6 m + h_2 {/eq}. Thus, the gravitational force on the person at sea level is {eq}F_{g1} = G \frac{m_p M_E}{r_1^2} {/eq}, and at an altitude of {eq}5000 m{/eq}, it is {eq}F_{g2} = G \frac{m_p M_E}{r_2^2} {/eq}, where {eq}m_p {/eq} is the mass of the person and {eq}M_E {/eq} is the mass of the Earth. The weight of the person is given by {eq}W = m_p g {/eq}, where {eq}g {/eq} is the acceleration due to gravity, which is \approx imately constant near the Earth's surface. The change in weight of the person is {eq}Delta W = W_2 - W_1 = m_p g_2 - m_p g_1 = m_p\left(frac{G M_E}{r_2^2} - \frac{G M_E}{r_1^2} right) {/eq}. Substituting the values, we get: {eq}Delta W = 100 kg\left(frac{6.674 times 10^{-11} mathrm{N} mathrm{m}^2/mathrm{kg}^2 times 5.972 times 10^{24} mathrm{kg}}{(6.38 times 10^6 m + 5000 m)^2} - \frac{6.674 times 10^{-11} mathrm{N} mathrm{m}^2/mathrm{kg}^2 times 5.972 times 10^{24} mathrm{kg}}{(6.38 times 10^6 m)^2} right) {/eq} Calculating this expression gives {eq}Delta W \approx 1.6 N{/eq}, which is option (c). Therefore, the weight of a person decreases slightly with increasing altitude.