What is the area of the SSME nozzle knowing only thrust at sea level and in vacuum?

Question

Looking at the SSME Wikipedia RS-25 Space Shuttle Engines, the data shown that the thrust at sea level ($F_{sl}$) is 1.816*10$^6$ N, while the thrust in vacuum ($F_{vac}$) is 2.278*10$^6$ N. Knowing only these two pieces of information, I need to find out the exit plane area of the Nozzle.

I understand $\dot{m}_{sl} = \dot{m}_{vac}$, but I cannot find the correct relationship to include the formula of Thrust $F_{*} = \rho*A_e*v^2_{e}$.

I found this question in the book I am currently studying. It is problem 1.2 of "Astronautics. The Physics of Space Flight", Second Edition, Ulrich Walter.

Answer 1

You can use the method I explained at the start of this answer: Falcon 9 Merlin 1d thrust calculated through every moment of flight

Quoting from that, you use the thrust equation:

$ \ \ \ F = \dot{m}_\mathrm e V_\mathrm e + (p_\mathrm e - p_0) A_\mathrm e$

You actually know four things, not two. You know the sea level thrust, the sea level ambient pressure, the vacuum thrust, and the vacuum ambient pressure.

Fill in those values and subtract the two equations from each other. The things you don't know (m-dot, Ve, and pe) drop out.

That gives you 462000 N = 101325 N/m^2 * Ae which resolves to

Ae = 4.55 m^2

which is close to the true area. (This fact sheet gives the exit diameter as 2.3 m)