Question

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
(kGM/R)^(1/2)
for k =

Answer 1

find the gravity's equation
and then use this-> mgh=1/2 mv^2

at top position it will have only PE
and when it will come to surface all the PE will convert into KE

its not alright tho..because there is a certain velocity after which the velocity will stop increasing(terminal velocity) <-we don't consider this because because no drag forces are specified
+
the gravity is not same everywhere but we assume it to be constant

Answer 2

really? I don't recall them teaching us this form of potential energy in class 9. weird.

Answer 3

they taught us tho
you talking about mgh ?

Answer 4

no, they taught that but I think it's better to use -GMm/R here

Answer 5

oh okay :)

Answer 6

yeah I wonder where she got this problem

Answer 7

@ParthKohli I got this problem from here- http://www.cracksat.net/sat2/physics/test603.html

Answer 8

And yeah/...9th problems are damn simple and boring...

Answer 9

take it to be 3R because we consider distance from the center

Answer 10

and well when it will strike the ground we must also consider mgR as PE

Answer 11

well then you'll need to know either of these:\[g=-\frac{GM}{R^2}\]\[\text{GPE} = \frac{-GMm}{R}\]

Answer 12

yeah..i took 3R and i am getting the wrong answer...please ek baar tum calculate karo na aayush...

Answer 13

yeah parth..i did that...

Answer 14

sorry i got 4/9* and correct is 4/3(that is what the site says)

Answer 15

\(g=\frac{-GM}{R^2}\)

\(mg(3R)=\frac{1}{2}mv^2 + mgR\)
\(mg(2R)= \frac{1}{2}mv^2\)
\( \frac{-GM}{R^2} (2R)= \frac{1}{2} v^2\)
\(\left(\frac{-4GM}{R} \right)^{\frac{1}{2}}=v\)

Answer 16

4/3 :/

Answer 17

\[\underbrace{\frac{1}2 mv^2}_{\text{increase in kinetic energy}} = \underbrace{GMm\left(\frac{1}R - \frac{1}{3R}\right)}_{\text{decrease in potential energy}}\]

Answer 18

okay..I was taking 3R in g ka formula and 2R for h in mgh...I guess that is where I erred...

Answer 19

But aayush...while finding g,don't we take center to center distance?

Answer 20

okay we had to take different gravities
\(g_{at~top}=\frac{-GM}{(3R)^2}\)
\(g_{at~surface}=\frac{-GM}{(R)^2}\)

\(mg_{top}(3R)=\frac{1}{2}mv^2 + mg_{surface} R\)
\(m \frac{-GM}{(3R)^2}(3R)-m\frac{-GM}{(R)^2}R= \frac{1}{2}mv^2\)
coming to same equations as pk's

Answer 21

stfu

Answer 22

yes

Answer 23

different gravities?