Determine the approximate value of g, at height D above the earth's surface, where D is the diameter of the earth.

Question

anonymous · Answer

Ok.. can you start somewhere?

anonymous · Answer

Vorbit $$=\sqrt{rg}$$

anonymous · Answer

g$$= (V _{orbit})^{2}/r$$

anonymous · Answer

$$\frac{ 7900^{2} }{ 6.37 \times 10^{6}+h }$$

anonymous · Answer

question is, what is h? is it 3 x r or 2x r. i would like to think that it is 2xr since the question says earth's surface, not  earth's atmosphere, or is it the same thing?

unknownrandom · Answer

I'm kind of thinking 2xr also. What level physics is this?

anonymous · Answer

first year university, A LEVEL and or second year in college

unknownrandom · Answer

Okay, my input might not be valuable considering I'm in a high school physics class.

anonymous · Answer

all is valuable. it's high school final year stuff. ok , if it is D  from the surface of the earth, then , THE Denominator or radius should be 2r.

anonymous · Answer

@Potatoes.ramu

anonymous · Answer

h is 2 times the radius = Diameter.. which is what is mentioned in your question!

anonymous · Answer

i thought as much, but for some reason, i got confused. but thank you @Mashy

prasadrajupvmv · Answer

g'=g(1-(2h/d))

anonymous · Answer

Wooow @prasadrajupvmv . am dead  lost!

prasadrajupvmv · Answer

what happend ? @MERTICH

anonymous · Answer

g'?

prasadrajupvmv · Answer

the new g value at height h