Answer: v₂ = 5962 km
the spacecraft will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits
Explanation:
Given that;
Lunar surface is in an altitude h = 43.0 km = 43 × 10³ m
we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²
speed of the space craft when it crashes into the lunar surface , v
decreasing speed of the space craft = 23 m/s
Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r
but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²
HERE r = Rm + h
we substitute
r = 1.74 × 10⁶ m + 43 × 10³ m
= 1.783 × 10⁶ m
On equating these, we have
G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)
v²/r = GM/r²
v = √ ( GM/r)
v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )
v = √ (2750787.9978)
v = 1658.55 m/s
Now since speed is decreasing by 23 m/s
the speed of the space craft into the lunar face is,
v₁ = 1658.55 m/s - 23 m/s
v₁ = 1635.55 m/s
Now applying conversation of energy, we say
1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)
v₂ = √ [ v₁² + GMe (1/Rm - 1/r)]
v₂ = √ [ 1635.55² + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]
v₂ = √ (2675023.8025 + 67979.24)
v₂ = √(2743003.046)
v₂ = 1656.2 m/s
now convert
v₂ = 1656.2 × 1km/1000m × 3600s/1hrs
v₂ = 5962 km
Therefore the spacecraft will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits
