IIT JEE , AIEEE Forum - Physics , Mechanics

sudakar

a planet revolves around the sun in an elliptical orbit of semi major axis 2*10¹²m. the areal velocity of the planet when it is nearest to the sun is 4.4*10¹⁶m²/s . the least distance between the planet and the sun is 1.8*10¹²m.then the min speed of the planet is in km/s is

sudakar

please someone reply

elastiboysai

Areal velocity dA/dt= L/2m ;L=angular momentum

i.e v*r/2=4.4*10^16

v*r=8.88*16

lemme call da nearest dist r1, farthest as r2, veloc max as v1, min. veloc as v2

now using geometry

a^2=b^2+c^2

a+c=5.56*10^11

now applyin angular momentum conservation

v1*r1=v2*r2

8.8*10^16=v2*5.56*10^11

wich is 1.582*10^5 m/s

sudakar

but the ans is 40000m/s and plzzz tell me how u got the first step

elastiboysai

well sorry sudakar i took semi minor axis to be2*10^12. sorry

any way repeat the same steps as i have done wid da correction

but still i don think da ans is 40000.

im gettin 23157 m/s

areal velocity is dA/dT= L/2m is a standard property, whose proof will be gvn in any standard text

sudakar

hey this a question from my reso test series and they have given me the ans as 40km/s. plzzz help

elastiboysai

hey yaar its rite tha ans is 40 km/s

again n again i made a mistake insubst.

my logic is correct

if u still don understand i can xplain it

ramyani

@ elastiboy,

will u plz du it a bit more elaborately?
plz explain the steps

a^2=b^2+c^2

a+c=5.56*10^11

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