The gain in kinetic energy if a 400-kilogram satellite moves from a distance of 3 × 106 meters above the surface of the Earth to a point 1.50 × 106 meters above the surface is _______ J. The mass of the Earth is 5.98 × 1024 kilograms and the radius of the Earth is 6.37x106 meters.
2006-08-21 15:40:27 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Astronomy & Space
how rude! and that coming from someone called "buttcheeks"! imagine that!
2006-08-21 15:49:57 · update #1
I love you Pancha...
D = Decrease
I = Increase
S_A Satallite at lower altitude where kin E 1/2 mv^2
hA = higher altitude
And 3.14~~ orbit (360 Degree orbit)
......... ...... .... ....... ..= GMm/r^2 = mv^2/r
so...... ... . . . . . .......= 1/2 mv^2
and.... (1/r2 - 1/r1)
because 1/1 = 1 + r =r1 = /\ (change) in Kin. E
because 1/1 = 1 + r = 1r
.........and the switch of the formula, from 1r and r1... so now it is 1/1 = 1 + r = r1
Now to check your answer.... on both sides by kg. and radii.
and...
<3.14~~/hA> (higher altitude) = GMm/r^ = mv^2/r =
........ . . . . . . . . . . . . . . . . .. = 1/2 mv^2
because (1/r^2 - 1/r1) because 1/1 = 1 + (r) = r1
thus: r1 /\ (change) in Kin. E
---AND---
{cos r^2/m/sin r1/kg.} *
. . . . .. . . . . . . .. . . . ... ../.........../
. . . . . . . . . . . ..5.98/x / + ~~ - / . .(x/1024 - 106/6.37)
.......... . . . . . . . .. . ...../. . . .. . ./
and...
... . . . . . . . . . . . . . . ../........../
{cos r^2/m/sin r1/kg.} /+ ~~ - / (6.37/1024 + 106/r1x^2)
. . . . .. . . . . .. . . . . .../........../
Fill in /\ (the change) in Kin. E apt. after completion of formula
The x variable is multiplied by itself as r1x^2 & thus:
cos r^2/m * sin r1/kg. = /\ (the change) in Kin. E
Furthermore, 6.37/1027 * sin r1/106 = cos of 6.37 as 1/1
because 106 of sin r1 - 1/1
and 1/1 = 1 or r/r as "cos" = r/r = r
thus: 1r switched is r1 =
.......... . . . and..... r1 = /\ (the change) in Kin E.
So, now the radii can be moved accordingly...
Hope this helped....
--Rob :)
2006-08-21 22:22:04 · answer #1 · answered by stealth_n700ms 4 · 0⤊ 0⤋
The satellite is going to a lower orbit, so its kinetic energy 1/2 mv^2 will increase, and its potential energy -GMm/r will decrease. For circular orbital motion, Newton's second law gives GMm/r^2 = mv^2/r, so 1/2 mv^2 = GMm/2r.
From that, you can calculate the change in kinetic energy (1/r2 - 1/r1) as you change orbital radii.
2006-08-22 02:04:06 · answer #2 · answered by bpiguy 7 · 1⤊ 0⤋
Pay attention in class and you wont have to make other people do your homework for you.
2006-08-21 22:45:37 · answer #3 · answered by buttcheeks 3 · 0⤊ 0⤋