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VladimirAG [237]

3 years ago

13

A 0.200 kg block on a vertical spring with spring constant of 4.65 ✕ 103 N/m is pushed downward, compressing the spring 0.084 m.

When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?

1 answer:

vlada-n [284]3 years ago

4 0

Answer:

PEel = PEg

PEel = .5 * k * x^2

= .5 * 4650 * .1^2

= 23.25

PEg= m * g * h

=.2 * 9.81 * h

23.25 = 1.962* h

h = 11.85

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Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

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$\epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2$

Value of specific energy is also given as

$\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km$

Orbit formula is given as

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Putting values in this equation and solving for e via the quadratic formula gives

$ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69$

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

$r_p=a(e-1)\\$

Substituting values gives

$r_p=133319 (1.086-1)\\r_p=11465.4 km$

Now for estimation of altitude z above earth is given as

$z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km$

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

$r_p=\frac{h^2}{\mu}\frac{1}{1+e}$

Rearranging this equation gives

$h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s$

Now the velocity at perigee is given as

$v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\$

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

$\delta =2 sin^{-1} (\frac{1}{e})$

Substituting value in the equation gives

$\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08$

Aiming radius is given as

$\Delta =a \sqrt{e^2-1}$

Substituting value in the equation gives

$\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km$

So the turn angle is 134.08 while the aiming radius is 5641.28 km

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