the acceralation due to gravity at a place is 0.2 m/s2 find. height
Answers
Answered by
2
use
g'=g/(1+h/r)^2
where g' is acceleration due to gravity in h km height. and r is the radius of earth
=> 0.2=10/(1+h/r)^2 (let g=10m/sec^2
=>(1+h/r)^2=50
=> 1+h /r=5root2
=> h=6r
g'=g/(1+h/r)^2
where g' is acceleration due to gravity in h km height. and r is the radius of earth
=> 0.2=10/(1+h/r)^2 (let g=10m/sec^2
=>(1+h/r)^2=50
=> 1+h /r=5root2
=> h=6r
abhi178:
oky
this. particulr problem is from. hc. verma 10 class. i am facing difficulty in physics. could ugive me some advice
you understand firstly concept then solve example problem. if you feel concetual strength then you solve their exercise . solve all question properly which is simple or hard . I think way your physics is strong. and you will enjoy this subject.
which. books. r good
hc verma ,if your English language strong read resnick heliday book, also if your target jee main and advanced read dc pandey
if you want deep knowledge of physics read cinage published B m sharma book
Thank you. then for. chemistry. and. maths
chemistry is enjoying subject in science read understand this by ncert . in physical chemistry RC mukherjee is the best writer. inorganic is memorable subject so for this read only ncert .. math iin higher class is calculus based so, my suggestion you deeply learn calculus .book for math is TMH
Thank. you. very. much
it's my pleaser
Answered by
1
The acceleration due to gravity is given by=gravitational force upon any object/
mass of an object
g=F/m=(GmM/r²)/m=GM/r²
where M=mass of earth,
r= gaps between the centre of the earth and object and all other symbols have their usual meaning.
0.2=(6.62×10^-11)×(6×10^24)/r²
r= 44732538.49 meter
mass of an object
g=F/m=(GmM/r²)/m=GM/r²
where M=mass of earth,
r= gaps between the centre of the earth and object and all other symbols have their usual meaning.
0.2=(6.62×10^-11)×(6×10^24)/r²
r= 44732538.49 meter
nice. but. abhi. is correct
Similar questions