Answer:
The shuttle's speed at the end of this stage
[tex]v_s = 2079.4 \ m/s [/tex]
The thrust
[tex]T = 2.5*10^{7} \ N [/tex]
The comparison shows that
[tex]T > W[/tex]
Explanation:
From the question we are told that
The initial mass is [tex]m_i = 2*10^{6} \ kg[/tex]
The final mass is [tex]m_f = 1*10^{6} \ kg[/tex]
The average speed is [tex]v = 3000 \ m/s[/tex]
The initial velocity is [tex]u = 0 \ m/s[/tex]
Generally the shuttle's speed at the end of this stage is mathematically represented as
[tex]v_s = v * ln [ \frac{m_i}{m_f} ][/tex]
=> [tex]v_s = 3000 * ln [ \frac{ 2*10^{6} }{ 1*10^{6} } ][/tex]
=> [tex]v_s = 2079.4 \ m/s [/tex]
Generally the the initial total weight of the shuttle is mathematically represented as
[tex]W = m_i * g[/tex]
=> [tex]W = 2*10^{6} * 9.8 [/tex]
=> [tex]W = 1.96*10^7 [/tex]
Generally the thrust during the same period is mathematically represented as
[tex]T = v * \frac{m_f}{m_i}[/tex]
=> [tex]T = 3000 * \frac{1*10^6}{2*10^{6}}[/tex]
=> [tex]T = 2.5*10^{7} \ N [/tex]
From our calculation we see that
[tex]T > W[/tex]
