Moon Gravitational Force Problem

AI Thread Summary
The discussion centers on calculating the frictional force acting on a lunar exploration vehicle with a mass of 5.68 × 10^3 kg, which requires a drive force of 1.42 × 10^3 N to achieve an acceleration of 0.220 m/s² on the moon. Participants clarify that the mass remains constant between Earth and the moon, and the frictional force is considered the opposite of the driving force. The conversation emphasizes the application of Newton's Second Law rather than gravitational laws for solving the problem. Ultimately, the participants confirm their understanding of the relationship between force, friction, and acceleration. The discussion concludes with a positive acknowledgment of the solution approach.
fewill
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1. In tests on Earth a lunar surface exploration vehicle (mass = 5.68 × 103 kg) achieves a forward acceleration of 0.220 m/s2. To achieve this same acceleration on the moon, the vehicle's engines must produce a drive force of 1.42 × 103 N. What is the magnitude of the frictional force that acts on the vehicle on the moon?
G= 6.673*10-11
mEarth= 5.98*1024
mMoon= 7.35*1022
rEarth= 6.38*106
rMoon= 1.74*106

2. F=Gm1m2/r2


3. All I know is that the mass is the same on Earth as it is on the moon. I figured that the frictional force is the opposite of the driving force. I am lost. Please help.
 
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fewill said:
1. In tests on Earth a lunar surface exploration vehicle (mass = 5.68 × 103 kg) achieves a forward acceleration of 0.220 m/s2. To achieve this same acceleration on the moon, the vehicle's engines must produce a drive force of 1.42 × 103 N. What is the magnitude of the frictional force that acts on the vehicle on the moon?
G= 6.673*10-11
mEarth= 5.98*1024
mMoon= 7.35*1022
rEarth= 6.38*106
rMoon= 1.74*106

2. F=Gm1m2/r2


3. All I know is that the mass is the same on Earth as it is on the moon. I figured that the frictional force is the opposite of the driving force. I am lost. Please help.

I'll give you a hint: this question has nothing to do with Newton's law of gravitation, but everything to do with Newton's Second Law :wink:
 
So do I use F-fk=ma?
 
fewill said:
So do I use F-fk=ma?
Sounds good to me :approve:
 
Got it thanks!
 
fewill said:
Got it thanks!
A pleasure.
 
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