- #1

i_hate_math

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## Homework Statement

Two small spaceships, each with mass m = 2000 kg, are in the circular Earth orbit of the figure, at an altitude h of 440 km. Igor, the commander of one of the ships, arrives at any fixed point in the orbit 81.0 s ahead of Picard, the commander of the other ship. What are the (a) period T0 and (b) speed v0 of the ships? At point P in the figure, Picard fires an instantaneous burst in the forward direction, reducing his ship's speed by 2.00%. After this burst, he follows the elliptical orbit shown dashed in the figure. What are the (c) kinetic energy and (d) potential energy of his ship immediately after the burst? In Picard’s new elliptical orbit, what are (e) the total energy E, (f) the semimajor axis a, and (g) the orbital period T? (h) How much earlier than Igor will Picard return to P?

## Homework Equations

kepler's 3rd law: T^2=4*(π^2)*(r^3)/(GM)

Newton's law: F=GMm/R^2

Centripetal Force: F=mv^2/R

1st Escape Speed= v=sqrt(GM/R)

KE=0.5mv^2=GMm/2R ------- by substitute v into the equation

GPE=-GMm/R

E=KE+GPE=-KE

where M=5.972*10^24 kg and m=2000kg

## The Attempt at a Solution

For (a), I used keeper's law T^2=4*(π^2)*(r^3)/(GM) and took the sqrt of RHS, got T=91.88s

I got this one wrong

b) i used mv^2/R=GMm/R^2

and got the expression of v=sqrt(GM/R) ----- or just take the escape speed

v=3.01*10^4 again WRONG

c)KE=0.5m(v')^2=8.69*10^11 ----- where v'=0.98v

again this is WRONG solution

d)GPE=-2KE or GPE=-GMm/R

sub in values got GPE=-1.74*10^12

e)E=KE+GPE=-KE=-8.69*10^11

now the rest depends on the correct solutions of previous ones, I wouldn't bother typing in the rest of my attempt. Please enlighten me with your genius minds