The radius of Venus (from the center to just above the atmosphere) is 6050 km (6

Question

The radius of Venus (from the center to just above the atmosphere) is 6050 km (6050103 m), and its mass is 4.91024 kg. An object is launched straight up from just above the atmosphere of Venus.

(a) What initial speed is needed so that when the object is far from Venus its final speed is 3000 m/s?
vescape =        m/s?
(b) What initial speed is needed so that when the object is far from Venus its final speed is 0 m/s? (This is called the "escape speed.")
vescape = 10417.72  m/s

I got part B on this problem, but dont know how to do part A? Thank you!!!

Explanation / Answer

Let M be the mass of Venus, m that of the object, r the distance from the center of mass of the planet to the object .
The force of attraction acting by the planet on the object is given by
F = - GMm/r²
The work of that force when the object escapes from distance ri = 6050×10^3 m to rf = infinity is the integral of F(r).dr taken on interval (ri, rf)
W = F(r)dr =-GMm(1/r²).dr = GMm(1/rf - 1/ri) = -GMm(1/ri)
. . (from ri to rf = inf.)
That work results in a change in kinetic energy equal to W
(Ek) = W = -GMm/ri
(1/2)m vf² - (1/2)m vi² = -GMm/ri
vf² - vi² = -GM/ri
vi² = vf² + GM/ri
In which
vf = 3000 m/s
G = 6.674×10^-11 N(m/kg)²
M = 4.9×10^24 kg
ri = 6.05×10^6 m

vi² = vf² + GM/ri
vi² = (3000)² + (6.674×10^-11) (4.9×10^24) / 6.05×10^6
vi² = 9000000 + (6.674×4.9 / 6.05)×10^7
vi² = 9000000 + 54×10^6 = 63×10^6
vi = 7937 m/s = 7.937 km/s

The initial speed must be 7.937km/s so that when the object is far from Venus its final speed is 3000 m/s

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