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.

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

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

+15-35=-20

going in opposite directions