Question

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 earth 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.

Answer 1

v₀ = 13.9 10³ m / s

Step-by-step 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