whatwill be theaaceleration due to gravity on a body at the heigh 4000km above the surface of earth
Answers
Answer:
3.675
Explanation:
gh=g(1−h/R)
where R
is radius of earth and it's value is equal to 6400KM.
Now,
gh=9.8(1-4000/6400)
3.675m/s*s (Approx)
Answer:
3.7 m/s^2
Explanation:
To solve for g at an altitude of 4000000 m from the earth surface, we need to know what does g=9.8 m/s^2 signifies.
The value g=9.8 m/s^2 is the value on the surface of earth at the sea level from the centre of earth.
Some useful numbers which will help in calculation,
Radius of earth = 6.38 x 10^6 meter.
Universal gravitational constant, G = 6.673 x 10^(-11) N-m^2/kg^2.
Mass of earth,M = 5.98 x 10^24 kg
(Don't worry you don't have to remember these values in exam since examiner will provide the above mentioned values for such questions.)
Now formulas to be used are,
(1) Force of gravity on an object of mass "m" , F = (m) x (g)
(2) F = { (G) x (M) x (m) } / (r^2)
equating both equations we get,
g = (G x M)/ r^2 ••••• (i)
Now the object is at a distance of 4000000 m from earth surface, therefore the value of
r = (6.38 x 10^6) + (4000000)
= 10,380,000 m
Putting the value of G, M and r in equation (i) we get
g = 3.7 m/s^2
ALTERNATE ANSWER
We can also use a short formula for such question as it will provide answer way quicker i.e.,
value of "g" at a height "h" = g {1 - (h/r)}
where r = 6.4 x 10^6 or 6400 km.