Question

What would be the gravitational field strength 2000km above the Earth? 50km above the Earth?

Answer 1

Gravity decreases with altitude, since greater altitude means greater distance from the Earth's center.

so the gravity at 2000 km above the earth surface will be very less as compare to 50 km above the earth surface.

Answer 2

\[F=GmM/(R _{e}+r)^2\]
where M is mass of the Earth & Re is radius of the Earth

Answer 3

what is Gm and r?

Answer 4

r is distance of mass to Earth and G is gravitation constant that equal to 6.63*10^34(i'm not sure about G's value but i think that.many of physics book has this eg holliday)

Answer 5

G is the gravitational constant and m is the mass of any object on the earth surface
it is not r it is h which means

Answer 6

i think hosein it is h because the question stats that is about the altitude

Answer 7

uhm, I don't really know the numbers to plug into the equation

Answer 8

yea here r is h

Answer 9

you can't solve this,bath?

Answer 10

so u can put 2000km as h or 5o as h
but u cannot solve it

Answer 11

ya

Answer 12

i don't know what m is

Answer 13

m is mass of 2nd body

Answer 14

m is the mass of a any object on the earth surface
or the object above the earth surface e.g a satellite

Answer 15

do you know what value I would plug into it?

Answer 16

mass of an object e.g 40 kg
but is negligible as compared to earths mass

Answer 17

G=6.67*10(-11) N(m/kg)^2
M = 5.9722 × 1024 kg

Answer 18

subtitute the values given if u have any word problem

Answer 19

you ought be consider that in formula

Answer 20

ok so would the equation then become \[G= 6.67\times10^{-11}\] \[M= 5.98 \times10^{24}\] \[R e = 6.38 \times10^{6 }\] \[h= (2000)or (50)\]

Answer 21

yea put it in formula

Answer 22

you should these standard values in the formula

Answer 23

and then \[((6.67\times10^{-11}) \times 40) \div ((6.38 \times 10^{6}) + 2000)\]

Answer 24

where is Earth mass? and denominator has power 2

Answer 25

but accelaration due to gravity too comes in the formula

Answer 26

we forgot the main thing

Answer 27

what thing?sgholap100

Answer 28

accelaration due to gravity

Answer 29

ican't understand your goal?can you explane

Answer 30

you can use gRe^2 instead of GMe

Answer 31

as the question say that the object is above earth surface
accelaration due to gravity(g=9.8m/s^2) too comes
then the formula becomes

Answer 32

your goal is?\[mg=GMm/R _{e}^2\]
am i right?

Answer 33

\[((6.67 \times 10 ^{-11}) \times 40) \times 5.98 \times 10 ^{24}) \div(6.38 \times 10 ^{6} +2000)^{2} \]

Answer 34

oh yeah i am sorry its was my mistake
i was considering
Effect of altitude on gravity
@hosein u r right

Answer 35

ithink it's right.

Answer 36

yes you r right

Answer 37

so would the answer be 391.7 ?

Answer 38

ok ya I think I got it, thanks guys :)

Answer 39

welcome
but do one thing neglect the m and h in the formula as they are very small as compared to M and R
OK
^_^