A 1.14 X 10"-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is t.60 rn/s1
-
At an altitude of 165 m, the craft's downward velocity is 18.0 rn/s. To
slow d~wn the craft, a retrorocket is firing to provide an upward thrust,
Assuming the descent is vertical, find the magnitude of the thrUst needed
to reduce the velocity to zero at the instant when the craft touches the
lunar surface.
Please proof read. I have no idea what the mass is and am guessing you mean g = about 9.81/6 = 1.64
positive x up
F up = m a =T - m g
so a up =T/m - g
v = Vi + a t
Vi = - 18
at surface v = 0
0 = -18 + a t
a t = 18
h = Hi + Vi t + (1/2) a t^2
0 = 165 - 18 t +(1/2) (18/t) t^2
9 t - 18 t + 165 = 0
9 t = 165
t = 18.3 seconds to stop
a = 18/t = 0.984 m/s^2
now back to
a =T/m - g
0.984 = (T/ whatever mass is) - whatever g is
To find the magnitude of the thrust needed to reduce the velocity to zero at the instant the craft touches the lunar surface, we can use the principle of conservation of energy. The potential energy lost during the descent will be equal to the work done by the retrograde rocket.
The potential energy lost during the descent can be calculated using the formula:
Potential energy lost = mass × acceleration due to gravity × height
Given:
Mass of the lunar landing craft = 1.14 × 10^3 kg
Acceleration due to gravity on the moon = 1.6 m/s^2
Height = 165 m
Potential energy lost = (1.14 × 10^3 kg) × (1.6 m/s^2) × (165 m)
Potential energy lost = 299,760 J
Now, the work done by the retrograde rocket will be equal to the potential energy lost. In order to reduce the velocity to zero, the work done should be in the opposite direction of the velocity. Hence, the work done by the retrograde rocket will be equal to the negative potential energy lost.
Work done by the retrograde rocket = -299,760 J
The work done can be calculated using the formula:
Work done = Force × distance
Since the distance is the height (165 m), the force exerted by the retrograde rocket can be calculated as:
Force = Work done / distance
Force = -299,760 J / 165 m
Force = -1818.91 N
The magnitude of the thrust needed to reduce the velocity to zero is approximately 1818.91 N.
To solve this problem, we need to use the principles of Newton's second law and the concept of work done by a force.
First, let's determine the initial velocity of the craft. We are given that the downward velocity is 18.0 m/s. Since the craft is descending, we can consider this as the negative velocity. So, the initial velocity (v_i) is -18.0 m/s.
Next, let's determine the final velocity when the craft touches the lunar surface. Since we want the craft to come to a complete stop (velocity = 0), the final velocity (v_f) is 0 m/s.
Using the kinematic equation:
v_f^2 = v_i^2 + 2ad
where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and d is the distance covered.
Since the craft is decelerating, the acceleration is in the opposite direction to the velocity. Therefore, we can take the acceleration (a) as -9.8 m/s^2 (approximate value of acceleration due to gravity on the moon).
The distance covered (d) is given as 165 m.
Substituting these values into the equation, we can solve for the final velocity:
0 = (-18.0)^2 + 2(-9.8)d
0 = 324 + (-19.6)d
19.6d = 324
d = 16.53 m
Therefore, the final velocity of the craft when it touches the lunar surface is 0 m/s and the distance covered is 16.53 m.
Now, let's calculate the work done by the retrorocket to slow down the craft. The work done is equal to the change in kinetic energy.
Using the work-energy principle:
Work done = Change in kinetic energy
Since the craft is coming to a stop, the change in kinetic energy is equal to the initial kinetic energy.
The initial kinetic energy (K_i) is given by:
K_i = (1/2)mv_i^2
where m is the mass and v_i is the initial velocity.
Given:
Mass (m) = 1.14 × 10^3 kg
Initial velocity (v_i) = -18.0 m/s
Substituting these values, we can find the initial kinetic energy:
K_i = (1/2)(1.14 × 10^3)(-18.0)^2
K_i = (1/2)(1.14 × 10^3)(324)
K_i = 186,120 J
Therefore, the work done by the retrorocket is 186,120 J.
Finally, let's calculate the magnitude of the thrust needed to reduce the velocity to zero. The force exerted by the retrorocket is equal to the work done divided by the distance covered.
Force = Work done / Distance covered
Force = 186,120 J / 16.53 m
Force ≈ 11,250 N
Therefore, the magnitude of the thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is approximately 11,250 N.