Asked by matin
A rocket is launched vertically from Earth's surface with a velocity of 3.4 km/s. How high does it go:
From Earth's centre?
From Earth's surface?
how can this be done without any time or force mentioned ??
From Earth's centre?
From Earth's surface?
how can this be done without any time or force mentioned ??
Answers
Answered by
Elena
vₒ= 3.4 km/s= 3400 m/s.
The problem may be solved by several methods.
The first method (kinematics)
The height from the Earth's surface
h =vₒ•t -g•t²/2,
v=vₒ-g•t.
At the top point v=0, => vₒ=g•t, t = vₒ/g,
Substitute it in the equation for h and obtain
h=vₒ²/2•g =(3400 )²/2•9.8 ≈ 5.9•10^5 m. =590 km.
The second method (the law of conservation of energy)
KE = ΔPE.
m•v²/2 = m•g•h,
h = vₒ²/2•g =….
The Earth’s radius is R ≈ 6378 km.
The distance from Earth's centre is
H = 6378 + 590 = 6968 km.
The problem may be solved by several methods.
The first method (kinematics)
The height from the Earth's surface
h =vₒ•t -g•t²/2,
v=vₒ-g•t.
At the top point v=0, => vₒ=g•t, t = vₒ/g,
Substitute it in the equation for h and obtain
h=vₒ²/2•g =(3400 )²/2•9.8 ≈ 5.9•10^5 m. =590 km.
The second method (the law of conservation of energy)
KE = ΔPE.
m•v²/2 = m•g•h,
h = vₒ²/2•g =….
The Earth’s radius is R ≈ 6378 km.
The distance from Earth's centre is
H = 6378 + 590 = 6968 km.
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