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adoni [48]
3 years ago
9

In order to rendezvous with an asteroid passing close to the earth, a spacecraft must be moving at 8.50×103m/s relative to the e

arth at a distance 2.50×108m from the center of the earth. At what speed must this spacecraft be launched from the earth's surface? Neglect air resistance and the gravitational pull of the moon.
Physics
1 answer:
Mariulka [41]3 years ago
6 0

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

    F = m a

    G m M / x² = m dv / dt = m dv/dx  dx/dt

    G M / x² = dv/dx   v

    GM dx / x² = v dv

We integrate

    v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m  and the velocity v = vo and the upper limit x = 2.50 10⁸m  with a velocity of v = 8.50 10³ m/s

    ½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

    72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

    72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3  10⁻⁸)

    72.25 10⁶ - v₀² = -1.213  10⁸

    v₀² = 72.25 10⁶ + 1,213 10⁸

    v₀² = 193.6 10⁶

    v₀ = 13.9 10³ m / s

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Tanzania [10]

Answer:

a)  r₁ = 3.836 10⁷ m,  b)   F = - 3,390 10⁸ N , c)  R = 120.3 m

Explanation:

a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.

       F₁ -F₂ = 0

       F₁ = F₂

       G m M_{m} / r₁² = G m M_{e} / r₂²

Let's look for the measured distance from a Coordinate System located on the Moon,

         r₂ = D - r₁

Where D is the average distance from Terra to the Moon 3.84 10⁸ m

        M_{m} / r₁² = M_{e} / (D - r₁)²

       (D² - 2 D r₁ + r₁²) = M_{e} /M_{m} r₁²

       (1 - M_{e} / M_{m})r₁² - 2D r₁ + D² = 0

Let's replace and solve the second degree equation

       (1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0

       -80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0

        r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0

        r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2

        r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2

The results are:

       r₁’= 38.355 10 6 m

       r₁ ’’ = -47.915 10 6 m

We take the positive result that a distance between the moon and the Earth, the equalization point is    r₁ = 3.836 10⁷ m

b) The force at point R = 2 r₁

        R = 2 3.8355 10⁷ = 7.671 10⁷ m

        F = F₁ - F₂

        F = G m  M_{m} / R² - G m  M_{e} / (D- R)²

        F = G m ( M_{m} / R² -  M_{e} / (D-R)²)

   F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²

        F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)

        F = m (-3.3897 10⁸) N

The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)

        F = - 3,390 10⁸ N

c) let's use gravitational force from the moon

        F = G m  M_{m} / R²

        R =√ G m  M_{m} / F

        R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)

        R = √ (1.4482 10⁴)

        R = 1.2034 10² m

        R = 120.3 m

8 0
3 years ago
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