Let us first determine the velocity of the descent module. We note that the change in pressure Δp is connected with the change in altitude Δh through the following relation:
where ρ is the gas density. The equation of state for an ideal gas implies that p = (ρ/μ) RT (here T is the gas temperature at the point where the change in pressure is considered). Taking into account that Δh = - v Δt, where v is the velocity of the descent, and Δt is the time of the descent, we can write expression (1) in the form
Knowing the ratio Δp/Δt, i.e. the slope of the tangent at the final point A of the graph, we can determine the velocity v from Eq. (2). (It should be noted that since the left-hand side of (2) constrains the ratio Δp/p, the scale on the ordinate axis is immaterial.) Having determined (Δp/Δt) p-1 from the graph and substituting μ = 44 g/mol for CO2, we find that the velocity of the descent module of the spacecraft is
Let us now solve the second part of the problem. Considering that the module has a velocity of 11.5 m/, it was at an altitude h = 15 km above the surface of the planet 1300 before landing, i.e. this moment corresponds to t = 2350 *. Using the relation (Δp/Δt) p-1, we can find the required temperature Th at this point of the graph from Eq. (2):