Answer:
[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]
Explanation:
For a lunar lander acceleration in an upward direction, the net force that acts on it can be expressed as:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]
The equation for the net force that acts on the lunar lander is:
[tex]F_{net} = m_La[/tex]
For the lander touch down to be zero acceleration, it implies that the acceleration of the moving rocket is zero(a free body fall)
i.e. a = 0
We can now regard the Apollo 12 lunar as a freely falling body
However; the force of gravity as a result of the moon acting on the lunar rocket is:
[tex]F_{g(moon)} = m_Lg_m[/tex]
Then; the equation for the thrust force of the lunar rocket is:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex]
[tex]m_La = F_{thrust}-m_Lg_m[/tex]
[tex]m_L(0)= F_{thrust}-m_Lg_m[/tex]
[tex]0= F_{thrust}-m_Lg_m[/tex]
[tex]-F_{thrust}= -m_Lg_m[/tex]
[tex]F_{thrust}= m_Lg_m[/tex]
Finally; the thrust force of the lunar rocket can be determined as:
[tex]F_{thrust}= m_Lg_m[/tex]
Acceleration due to gravity ot the surface of the moon = 1.625 m/s²
[tex]F_{thrust}=(1.4 \times 10^{4} \ kg) (1.625 \ m/s^2)[/tex]
[tex]F_{thrust}=2.275 \times 10^4 \ N[/tex]
[tex]\mathbf{F_{thrust} \simeq 2.3 \times 10^4 \ N}[/tex]
A thrust force of 22680N will be necessary to to have the lander touch down with zero acceleration.
Given the data in the question;
Mass of Apollo 12 lunar lander; [tex]m_L = 1.4 * 10^4 kg[/tex]
Using the expression for the net-force that acts on the lander when it is accelerated in the upward direction:
[tex]F_{net} = F_{thrust} - F_{g(moon)}[/tex] ------- Let this be equation 1
Also, the net-force that acts on the lunar lander can be expressed as:
[tex]F_{net} = m_L*a[/tex] ------- Let this be equation 2
Where [tex]m_L[/tex] is the mass of the lunar lander and a is the acceleration ( 0 ),
( the acceleration of the moving rocket is zero and its freely falling )
Also, Force of gravity due to moon that acts on the lunar lander is expressed as:
[tex]F_{g(moon)} = m_L * g_m[/tex] ------- Let this be equation 3
Where [tex]m_L[/tex] is the mass of the lunar lander and [tex]g_m[/tex] is the moon's gravity ( [tex]1.62m/s^2[/tex] )
Lets substitute equation 2 and 3 into 1
[tex]m_L * a = F_{thrust} - (m_L * g_m)\\\\m_L * 0= F_{thrust} - (m_L * g_m)\\\\0=F_{thrust} - (m_L * g_m)\\\\ F_{thrust} = m_L * g_m[/tex]
We substitute our values into the equation
[tex]F_{thrust} = (1.4*10^4kg) * 1.62m/s^2\\\\F_{thrust} = 22680 kg.m/s^2\\\\F_{thrust} = 22680N[/tex]
Therefore, a thrust force of 22680N will be necessary to to have the lander touch down with zero acceleration.
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