Force constant of a spring launching a satellite

In summary, to design a spring that will give a 1200 kg satellite a speed of 3.80 m/s relative to an orbiting space shuttle with a maximum acceleration of 5.00g, the force constant of the spring must be (25mg^2)/v^2 and the distance the spring must be compressed is (5mg)/k. This can be derived from the equations F=kx and 0.5mv^2 = 0.5kx^2, where F is the force, k is the force constant, x is the distance the spring is compressed, m is the mass of the satellite, g is the acceleration due to gravity, and v is the desired speed of the satellite.
  • #1
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You are asked to design a spring that will give a 1200 kg satellite a speed of 3.80 m/s relative to an orbiting space shuttle. Your spring is to give the satellite a maximum acceleration of 5.00g. The spring's mass, the recoil kinetic energy of the shuttle, and changes in gravitational potential energy will all be negligible.

a) What must the force constant of the spring be?

b) What distance must the spring be compressed?

I don't really know how to approach this problem, since without the spring constant (k), you cannot find the distance the spring is compressed (x) and vice versa.

But, according to the problem,

0.5kx^2 + mgh = 0.5mv^2 +5mgh
0.5kx^2 = 0.5mv^2 + 4mgh
kx^2 = mv^2 + 8mgh
k = (mv^2 + 8mgh)/x^2

I know this has to be wrong because there are three unknowns, h, x, and k. Can someone help me approach this differently?
 
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  • #2
You were told to ignore the gravitational potential energy, so drop the mgh terms. Then your energy equation is correct, but you still have two unknowns. Now use the condition that the maximum acceleration the spring is supposed to impart is 5g. (F=ma etc).
 
  • #3
0.5mv^2 = 0.5kx^2

I am not seeing how to incorporate acceleration into this equation. To find the work done by the force of the spring,

F= ma = 5mg * x = 5mgx ??
 
  • #4
For a spring, F=k*x, right? You don't incorporate it into that equation, you derive another equation. 2 equations+2 unknowns=happiness.
 
  • #5
ahhhh.

so

F=kx
5mg=kx
x=(5mg)/k

0.5mv^2 = 0.5kx^2
mv^2 = kx^2
mv^2 = k([5mg]/k)^2
mv^2 = k(25m^2g^2)/(k^2)
k = (25mg^2)/v^2

right? i think so, yay
 

1. What is the force constant of a spring?

The force constant of a spring, also known as its spring constant, is a measure of the stiffness of the spring. It is represented by the letter k and is defined as the force required to stretch or compress the spring by a unit of length.

2. How is the force constant of a spring related to launching a satellite?

The force constant of a spring is a crucial factor in determining the launch velocity and trajectory of a satellite. It is used in the calculation of the spring force, which is necessary to overcome the gravitational pull of the Earth and launch the satellite into orbit.

3. How is the force constant of a spring calculated?

The force constant of a spring can be calculated by dividing the force applied to the spring by the displacement it undergoes. It can also be determined by measuring the frequency of oscillation of the spring using the equation k = mω2, where m is the mass attached to the spring and ω is the angular frequency.

4. What factors can affect the force constant of a spring?

The force constant of a spring can be affected by factors such as the material and dimensions of the spring, as well as the temperature and any external forces acting on the spring. A change in any of these factors can alter the stiffness of the spring and therefore its force constant.

5. Why is the force constant of a spring important in satellite launches?

The force constant of a spring is important in satellite launches because it determines the amount of force needed to launch the satellite into orbit. A higher force constant means a stiffer spring, which requires more force to stretch or compress, resulting in a higher launch velocity. This is crucial for ensuring that the satellite reaches its desired orbit and stays in orbit for as long as required.

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