Respuesta :
Answer:
a) v = 1.79 10³ m / s, b) R_max = 1.33 10⁶ m , c) v = 1.26 10³ m / s , d)g = 3.2016 m / s²
Explanation:
a) to find the escape velocity let us use energy conservation
Initial. On the surface
Em₀ = K + U = ½ m v² - G m M / R
Final. Far from the asteroid
Emf = U = - G m M / R_max
Em₀ = Emf
½ m v² - G m M / R = - G mM / R_max
To escape the asteroid the distance must be infinite, so the last term is zero
v =√2GM / R
v = √ (2 6.67 10⁻¹¹ 1.2 10²²/500 10³)
v = √ 3.2 10⁶
v = 1.79 10³ m / s
b) how high can it go if it leaves with a speed of v = 1000 m / s
½ m v² - G m M / R = - G mM / R_max
1 / R_max = 1 / R - ½ v² / GM
1 / R_max = 1/500 10³ - ½ 1000² / (6.67 10⁻¹¹ 1.2 10²²)
1 / R_max = 2 10-6 - 1,249 10-6 = 0.751 10-6
R_max = 1.33 10⁶ m
c) The speed if the object falls from h = 1000 km above the superfine
The distance from the center of the asteroid is
R ’= R + h
R ’= 500 + 1000 = 1500 km
R ’= 1.5 106 m
½ m v² - G m M / R = - G mM / R ’
½ v² = GM (1 / R - 1 / R ’)
v = √ 2 GM (1 / R- 1 / R ’)
v = √ (2 6.67 10⁻¹¹ 1.2 10²² (1/5 10⁵ - 1/1 10⁶)
v = √ 1.6 10¹² (1 10⁻⁶)
v = 1.26 10³ m / s
d) the gravity acceleration of the asteroid is
g = G M / R²
g = 6.67 10⁻¹¹ 1.2 10²² / (5 10⁵)²
g = 3.2016 m / s²