The radius of Mercury (from the center to just above the atmosphere) is 2440 km

Question

The radius of Mercury (from the center to just above the atmosphere) is 2440 km (2440 × 103 m), and its mass is 3 × 1023 kg. An object is launched straight up from just above the atmosphere of Mercury.

(a) What initial speed is needed so that when the object is far from Mercury its final speed is 3000 m/s? Entry field with incorrect answer 4050 m/s

(b) What initial speed is needed so that when the object is far from Mercury its final speed is 0 m/s? (This is called the "escape speed.") Entry field with incorrect answer m/s

I am getting stuck on a) and need help with b). My initial answer for a was 4050

Explanation / Answer

a)
Look at the initial and final energies of the object. Far from Mars means zero gravitational potential energy.
Initial Energy=Initial Kinetic Energy + Gravitational Potential Energy
Final Energy=Final Kinetic Energy
(1/2)m1v1^2+(-Gm1m2/r)=(1/2)m1v2^2
v1^2=(2Gm2/r)+v2^2
v1^2=(2*6.67x10^(-11)(3x10^23)/2440 × 10^3 m)+3000 ^2
v1^2=25401639.34
v1=5040.009 m/s
.
.
.
b) Initial Energy=Initial Kinetic Energy + Gravitational Potential Energy
Final Energy=0
(1/2)m1v1^2+(-Gm1m2/r)=
v1^2=2Gm2/r
v1^2=2*6.67x10^(-11)(3x10^24)/2440*10^3
v1^2=164016393.4
v1=12806.888 m/s

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