Melodious
  • Melodious
The moon's mass is approximately 1% of the earth's mass. How does the gravitational pull of the earth on the moon compare with the gravitational pull of the moon on the earth?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Let Mass of Earth be Me Mass of Moon Mm Gravitational Pull of Earth Fe and Gravitational Pull of Moon Fm \[M_{e},M_{m},F_{e},F_{m}\] These are the quantities Now the gravitational pull is proportional to the mass of the body so \[F \propto M\] Removing the proportionality we introduce a constant k \[F=kM\] Now for earth and moon we have the equations \[F_{e}=kM_{e}\] \[F_{m}=kM_{m}\] Dividing these we get \[\frac{F_{e}}{F_{m}}=\frac{M_{e}}{M_{m}}\] From this we get \[F_{m}=\frac{M_{m}}{M_{e}}.F_{e}\] But you are also given Mm as 1% of Me so \[M_{m}=\frac{1}{100}.M_{e} \implies M_{e}=100M_{m}\] Substitute this value in equation for Fm and simplify
Melodious
  • Melodious
can explain this to me pls @Nishant_Garg

Looking for something else?

Not the answer you are looking for? Search for more explanations.