Question

A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station(radius of earth 6400km)

Answer 1

Whats wrong with my approach?
\[GMm/2r" ^{2}=GMm/r ^{2}\]
\[r=\sqrt{2(4096*10^{10})}\]
r=8960000
Distance from earth's surface=8960000-6400000m=2560km
The ans is supposed to be 2650km

Answer 2

Is r the earth's radius?

Answer 3

distance of separation of centre of earth and satellite when force is half its initial value

Answer 4

If r is the new distance, and R is the radius of the earth(i.e the initial separation) then:
\[r =\sqrt{2R}\]
Why have you substituted R=4096X10^10?
R=6400km

Answer 5

GMm cancel
so
sqrt(2r"^2)=r
?

Answer 6

F(R+h)=0.5F(R)
1/(R+h)^2= 0.5/R^2
2R^2=(R+h)^2

Answer 7

where exactly did i go wrong?

Answer 8

plz do it with my method and tell me what i did wrong

Answer 9

even with this method i get the same ans

Answer 10

give some time...
you didnt,i think so!!!
2R^2=(R+h)^2
81920000=40960000+12800h+h2
h2+12800h+40960000=0
check this!!!!

Answer 11

\[h=(\sqrt{2}-1)r\]
h=(.4)(6400)=2560

Answer 12

is this the solution to my quadratic ????

Answer 13

i think the answer is wrong,maybe!!!

Answer 14

this is from hc verma
i dont think hes ever wrong

Answer 15

no there are some errors in this book also, i too practised from his book sometimes...

Answer 16

Yes that is correct, shayan. The answer could be quadratic.
You can google H.C. Verma solutions to check this.
Use a calculator, Sarkar

Answer 17

mate which calc...wolf has a different solution

Answer 18

http://www.wolframalpha.com/input/?i=++find+the+values+of+h+in+h2%2B12800h%2B40960000%3D0

Answer 19

Quoting from HC Verma sol
h=0.414*6400=2649.6

Answer 20

but what exactly is wrong with the first method?

Answer 21

please give link to the solutions...

Answer 22

Must be a calculation mistake.

Answer 23

google search download
cant view

Answer 24

i agree @Mani_Jha

Answer 25

the solutions books takes a more accurate value of sqrt2
but what about my method what went wrong?

Answer 26

oh so i guess im supposed to take the more accurate value of root 2 as well

Answer 27

Whats wrong with my approach?
GMm/2r"2=GMm/r2
r=2(4096∗1010)−−−−−−−−−−−√
i think i got it!!
what is 4096*10^10 ????

Answer 28

it should be 6400exp3

Answer 29

where did root go?

Answer 30

r=R*sqr2
r=6400exp3* 1.4142

Answer 31

i got the ans by taking 1.414 for root2
but who in the holy world does that?

Answer 32

r=9050966m
R+h=9050966m
h=2650966m

Answer 33

everyone does that...

Answer 34

lol
i never do

Answer 35

r=sqrt(2(4096∗1010)−−−−−−−−−−−√
r=8960000
Check this with calculator

Answer 36

then you wont get an accurate answer,lol

Answer 37

no, i wont @Mani_Jha

Answer 38

thx people

Answer 39

There's a calculation mistake there.

Answer 40

@Mani_Jha -what are you typing,man -type fast....

Answer 41

Why? Got to catch a flight? :P

Answer 42

rather, gotta catch a bus!!!!

Answer 43

Have a safe journey!
Did u find your mistake, shayan?

Answer 44

yeah i did
i think i mentioned that