Ameteoroid, heading straight for earth, has a speed of 14.8 km/s relative to the center of earth as it crosses our moon's orbit, a distance of 3.84 × 108 m from the earth's center. what is the meteroid’s speed as it hits the earth? you can neglect the effects of the moon, earth’s atmosphere, and any motion of the earth. (g = 6.67 × 10 -11 n ∙ m2/kg2, mearth = 5.97 × 1024 kg)

v = 112.424 km / s

Explanation:

Given:

d = 3.84 x 10 ⁸ m  , R = 63 x 10³ m  , m earth = 5.97 x 10²⁴ kg  ,  G = 6.67 x 10⁻¹¹ N * m² / kg² , v₁ = 14.8 x 10 ³ m / s

Using the equation

KE + PE = initial KE + PE

¹/₂ * m * v₂² - G * m/ R = ¹/₂ * m * v₁² - G * m/(R+d)

v₂² = v₁² + 2*G * m [ 1/R - 1/(R+d) ]

v₂² = 14.8 x 10 ³ m /s + 2 * 6.67 x 10⁻¹¹ N * m² / kg² * 5.97 x 10²⁴ kg  [ 1 / 63 x 10³m - 1 / ( 63 x  10³ + 3.84 x 10⁸ m ) ]

v = √ 1.26 x 10¹⁰ m² / s²

v =  112.424 x 10³ m/s  

v = 112.424 km / s

would be false hope this out