Lunar Gravity velocity on impact

In summary, on the moon, the acceleration due to gravity is $-1.6m/s$. A stone dropped from a cliff on the moon and hits the surface of the moon $20s$ later. It falls a total of 320m, with a velocity of -32m/s at impact.
  • #1
karush
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Lunar Gravity, On the moon, the acceleration due to gravity is $-1.6m/s$ . A stone dropped from a cliff on the moon and hits the surface of the moon $20s$ later.
How far did it fall?
What is the velocity at impact.?

well here I presume since the rock is just dropped that $v_0=1.6m/s$ but the time is known t=20s and

$f(t)=-1.6t^2+1.6t+S_0$

$f'(t)=-3.2t+1.6

but I don't think this is set up right.
Ans 320m; -32m/s
 
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  • #2
Just a trivial matter perhaps, but the units of acceleration are distance per time squared.

Now, recall the kinematic equation for displacement given constant acceleration is:

$\displaystyle s(t)=\frac{1}{2}at^2+v_0t+s_0$

Since the rock is dropped, its initial velocity is zero. Its final height is zero, and it fell for 20 seconds. Hence, we may state:

$\displaystyle f(t)=-0.8t^2+s_0$

Solve for $\displaystyle s_0$ given:

$\displaystyle f(20)=0$

To find the velocity at impact, evaluate:

$\displaystyle f'(20)$
 
  • #3
karush said:
Lunar Gravity, On the moon, the acceleration due to gravity is $-1.6m/s$ . A stone dropped from a cliff on the moon and hits the surface of the moon $20s$ later.
How far did it fall?
What is the velocity at impact.?

well here I presume since the rock is just dropped that $v_0=1.6m/s$ but the time is known t=20s and

$f(t)=-1.6t^2+1.6t+S_0$

$f'(t)=-3.2t+1.6

but I don't think this is set up right.
Ans 320m; -32m/s

1. The units of the acceleration due to gravity are \(m/s^2\)

2. Dropped usually means that \(v_0=0\)

3. \( \displaystyle s(t)=\frac{g\; t^2}{2}+v_0 t +s_0 = - \frac{1.6\; t^2}{2}+s_0\)

When the rock hits the surface \(s(t)=0\), so ...

4. \(\displaystyle v(t)=g\;t+v_0 = \) ...

5. Since this is in calculus you need the supporting calculus: The equations in (3) and (4) above are obtained by integrating \(\ddot{s}= g\) and using \(v(t)=\dot{s}(t)\) with initial conditions \(\dot{s}(0)=v(0)=v_0\) and \(s(0)=s_0\).
 
Last edited:
  • #4
thanks much everyone, that was a great help:) $0=-0.8(20)^2+S_o\ \ S_o=320m$

$f'(20)=-1.6(20) =-32m/s$will be back with more, seem to slip and slide with this stuff.
 

Related to Lunar Gravity velocity on impact

1. What is lunar gravity velocity on impact?

Lunar gravity velocity on impact refers to the speed at which an object or spacecraft will hit the surface of the moon. This velocity is affected by the moon's gravitational pull and the mass and trajectory of the object.

2. How is lunar gravity velocity on impact calculated?

Lunar gravity velocity on impact is calculated using the formula V = √(GM/R), where V is the velocity, G is the gravitational constant, M is the mass of the moon, and R is the distance between the object and the center of the moon. This formula takes into account the gravitational force between the object and the moon.

3. What is the average lunar gravity velocity on impact?

The average lunar gravity velocity on impact is approximately 1.62 meters per second. This is much lower than the Earth's gravity velocity on impact, which is around 9.8 meters per second.

4. How does lunar gravity velocity on impact affect spacecraft landings?

Lunar gravity velocity on impact is a crucial factor in spacecraft landings on the moon. If the velocity is too high, the spacecraft could crash or sustain damage upon impact. If the velocity is too low, the spacecraft may not have enough momentum to reach the desired landing spot.

5. Can lunar gravity velocity on impact be manipulated?

Yes, lunar gravity velocity on impact can be manipulated through various means, such as using thrusters or adjusting the trajectory of the spacecraft. This is done to ensure a safe and precise landing on the moon's surface.

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