<p>SAT 2 physics
page 150, question 9:
The moon has mass M and radius R. A small object is dropped from a distance of 3R from the moon's center. The object's impact speed when it strikes the surface of the moon is equal to
square root of (kGM/R)</p>
<p>A) 1/3
B) 2/3
C) 3/4
D) 4/3
E) 3/2</p>
<p>I got k=4/9. Can anyone get this? How do you get the answer?</p>
<p>Answers and explanations are at the end of the book.</p>
<p>i think the answer in the book is wrong. How do you solve this question?</p>
<p>PE=KE+PE</p>
<p>-GMm/R=1/2mv^2-GMm/R (m’s cancel out)</p>
<p>-GM/3R=1/2v^2-GM/R
2GM/3R=1/2v^2
4GM/3R=v^2
v=square root (4GM/3R)
answer is D</p>
<p>why would the object still have potential energy after it strikes the surface of the moon?</p>
<p>Because for the Newton laws to work, the moon has to be treated as an infinitesimal point, despite the fact that its the planet. Thus a body on the surface of the moon is subject to gravitational force inversely proportional to R (radius of the moon) and consequently it has a potential energy, which is also inversely proportional to R. In other words, think about a tiny point at the center of the moon that is exerting the gravitational force on the body and is the cause of potential energy of the body.</p>
<p>why minus GM/3R in the equation -GM/3R=1/2v^2-GM/R
why not plus?
Isn’t the equation:
PE1+KE1=PE2+KE2
so why not:
GM/3R+0=GM/R+1/2v^2??</p>
<p>Why does the object have a lower potential energy(GM/3R) when it is 3R from the moon when it has more potential energy(GM/R) at the surface of the moon??</p>
<p>Okay sorry but what topic is this about anyway?! Like whats the chapter’s name what does it discuss?
I know dumb question but pleas see reply</p>