Question

(a) At what height above the earth is the acceleration due to gravity 44% of its value at the surface?


m

(b) What is the speed of a satellite orbiting at that height?
m/s

Answer 1

\[F_{grav} = G\frac{mM}{r^2}\]
where r is the distance from the CENTER of the earth.
Let R be the radius of the earth.
\[F_{grav,surface} = G\frac{mM}{R^2}\]
\[.44F_{grav,surface} = G\frac{44}{100}\frac{mM}{R^2}=G\frac{mM}{r^2}\]
Solve for r

Answer 2

what is mM

Answer 3

two masses. One is the mass of the earth, the other is the mass of the object. It doesn't matter which one you use for which, as long as you are consistent.

Answer 4

Dont u need the solution of ur question b)

Answer 5

this is what i have so far?

Answer 6

@Vandreigan

Answer 7

You won't need most of that stuff.

\[G(.44)\frac{mM}{R^2} = G\frac{mM}{r^2}\]
\[\frac{.44}{R^2}=\frac{1}{r^2}\]
\[r = \sqrt{\frac{100R^2}{44}}\]
\[r = \frac{5R}{\sqrt{11}} = \frac{5\sqrt{11}R}{11}\]

Answer 8

wouldn't that just come out to R=r

Answer 9

would R be 6778100 (the radius of earth in M)

Answer 10

Nope
\[r = \frac{5R}{\sqrt{11}}\]
You can just plug in the radius of the earth to find r.
Remember, though. This is the distance from the CENTER of the earth.

Answer 11

i get 10218370.2 (but i don't think thats right)

Answer 12

i only have like 10 minutes until my computer dies

Answer 13

R = 6371000m
Therefore, r = 9604643.882m

Answer 14

Distance above the ground is then:
r-R = 3233643.882

Answer 15

ooooohhh

Answer 16

so how do i find the speed

Answer 17

As for the speed of the satellite, we can just use:
\[a = \frac{v^2}{r}\]
Where a is the acceleration due to gravity at the distance r (44% of what it is on the surface)

Answer 18

6435.46?

Answer 19

\[v = \sqrt{.44*g*r}\]
g = 9.81 m/s^2
r = 9604643.882m

Answer 20

the answer was 435.46 thanks so much for your help! gotta go