Calculating Force & Energy of Impact on Deimos

[kepler]

Hi,

If an astronaut of 75 Kg of mass gives a jump of 1.5 meters in the air and hit the ground of the martian satellite Deimos then:

1) what will be the force of the impact in Newtons?
2) What will be the energy of the impact in Joules?
3) Will Deimos move?

I think we must first calculate g for Deimos. I have the following data regarding the mass and radius - from Wikipedia

g = G*2.244*10^15/6300^2 ~ 0,003771 m/s²

The website refers 0.0039 m/s², so the radius must be rounded in their data ( they refer the diameter as 12.6 km ) and G must have more decimal places.

From here, the impact of the astronaut would be, I think:

EK = 75 * 0.0039 * 1.5 = 0,43875 Joules

The force, I'm not sure. 75 kg is the mass of the astronaut. 75 * 0.0039 ~ 0,2925 is its weight. Since F=ma, I think that the force in Newtons of the impact is equal to its weight, or 0.2925 Newtons. Am I right so far?

So, Deimos will suffer an acceleration of

F = ma
0,2925 = 2,244*10^15 * a
a = 1,303476e-16 m/s²

( this part is confused...does it moves how much? )

Can someone help me out here? This is urgent.

Kind regards,

kepler

 

[OlderDan]

You cannot answer #1 with the information given. You would need to know how long it takes for the astronauts velocity to go to zero after first contacting the ground.

It looks like you are doing the right thing with #2, but I did not try to verify your calculations. The astronaut will have potential energy mgh at the highest point, and that energy must be absorbed on impact.

For #3, think about conservation of mometum. What must happen to the satellite if the astronaut pushes off to jump in the air? What must happen as the astronaut is coming back down?

 

[Dr.Brain]

'g' for Deimos is given by:

$g_D= \frac {GM}{R^2}$


Conservation of energy is (also) applicable on Deimos Surface , so :

$mgh= \frac {1}{2}mv^2$

For the first part , use $v^2 - u^2 = 2as$ , here u=0 and v can be calculated by using conservation of energy principle. Then force of impact = m'a'

For the second part, the lost of potential energy would be the energy of impact and is given by $mgh$

BJ

 

[kepler]

'g' for Deimos is given by:

$g_D= \frac {GM}{R^2}$


Conservation of energy is (also) applicable on Deimos Surface , so :

$mgh= \frac {1}{2}mv^2$

For the first part , use $v^2 - u^2 = 2as$ , here u=0 and v can be calculated by using conservation of energy principle. Then force of impact = m'a'

For the second part, the lost of potential energy would be the energy of impact and is given by $mgh$

BJ

Ok... so, the energy, in Joules, is well calculated, right? Since it's equal to mgh.

As for the force, I don't understand very well what you mean. I know the velocity is sqrt(3g) and the fall time is sqrt(3/g). So according to you

(sqrt(3g))^2-0=2ah

3g=2*a*1.5

g = a

So F = mg

Isn't it right? But it can't be. A body that falls from 1 meter has a force in impact different than if it falls from 2 meters high.

Kepler

 

[OlderDan]

For the first part , use $v^2 - u^2 = 2as$ , here u=0 and v can be calculated by using conservation of energy principle. Then force of impact = m'a'

BJ

And what are you going to use for s ?

You don't know how the astronaut lands. If he lands stiff legged on hard ground, s is zero and the force is infinite. If the ground is soft, or he bends his knees to absorb the shock, the force will be less. Since none of that information is given, you cannot calculate the force of impact.

 

[Dr.Brain]

And what are you going to use for s ?

You don't know how the astronaut lands. If he lands stiff legged on hard ground, s is zero and the force is infinite. If the ground is soft, or he bends his knees to absorb the shock, the force will be less. Since none of that information is given, you cannot calculate the force of impact.

If an astronaut of 75 Kg of mass gives a jump of 1.5 meters in the air and hit the ground of the martian satellite

I think we need not include "the mood of the astronaut" in physics problems.

BJ

 

[kepler]

...

Ok guys...

Forget it's an astronaut. Forget it's Deimos. Let's bring the problem "home".

If a body ( perfect ball ) with a mass ' m ' Kg falls from ' h ' meters high ( initial velocity = 0) hitting the Earth's ground, what will it be the Force of the Impact in Newtons? We know:

1) the velocity it has reaching the ground
2) the time the fall lasts
3) the kinetic energy
1) the momentum

What's the Force of Impact in Newtons? Oh, let's assume Earth has an hard ground. Is this better ? :)

Regards,

Kepler

 

[Dr.Brain]

Force of impact will be the same as in my first post.

Explaination for Energy Of Impact: When the body loses the potential energy , it acquires equal amount of kinetic energy, this is the energy with which it makes the impact.

BJ

 

[HallsofIvy]

As Doc Al has said several times, you cannot calculate the force based only on that information. You would also have to know how the object is stopped. You can calculate both kinetic energy and momentum of the object just before hitting the ground and, of course, both of those must become 0, but unless you know either the time duration of the impact or the distance moved during impact you cannot calculate the force. If the impact extends over a long time or distance, the acceleration, and so force, will be small. If the impact is sudden and short, the acceleration and force will be large.

 

[kepler]

Regarding the impact being equal to the kinetic energy ( m . g . h ), I'm not sure... if a body of 10 Kg falls from the height of 200 meters it will have the same KE of another body of 50 Kg falling from 40 meters. Which would you prefer to be hit with ? Since accelaration ' g ' is pratically the same at 40 or 200 meters high, and F = m . g, the first object will have a Force 5 times less than the second one even if its velocity is 5 times faster.

I agree with the Force being calculated by dividing ' m . (vf - vi) ' ( change in the momentum ) by the time of duration of the collision - vf is zero, and we know vi.

The time of collision will probably be less than a second - most collisions are, I think. So can we, or not, state that the impact force will be equal or greater than ' m . vi / 1 ' ? ( if the time is less that a second, 0.01 for example, the division will increase the value of F by 100 ).

On the other hand, when a car passes in the street where we leave, the windows of our home often shake for some seconds. If a friend of ours jumps in the living room, we sense the impact ( trembling ) for one or two seconds...

I don't know...maybe we must assume that the hiting ground is as hard as a rock so the formula be applyed.

Kind regards,

Kepler