The gravitational force will double as well because
F=mg
Answer:
As <em>Force </em><em>=</em><em> </em><em>Mass </em><em>×</em><em> </em><em>Acceleration.</em><em> </em><em>So </em><em>If </em><em>mass </em><em>is </em><em>constant,</em><em> </em><em>the </em><em>acceleration </em><em>would </em><em>also </em><em>be </em><em>constant,</em><em> </em><em>in </em><em>presence </em><em>of </em><em>a </em><em>net </em><em>external </em><em>constant </em><em>force!</em>
DIAMONDS are NOT rocks. Diamonds are minerals, rocks are made up of many different fragments of minerals.
Use energy conservation, since no energy is lost it must be constant.
E = 0.5mv² + mgh
At release the velocity v = 0 and the height is h.
E = 0 + mgh
At impact the height h = 0 and the velocity is v.
E = 0.5mv² + 0
Since the energy E is conserved:
0.5mv² = mgh
the mass m cancels and the equation becomes:
0.5v² = gh
h = 0.5v²/g
when g = 9.81 and v = 22:
h = 24,66
m = mass of the person = 80 kg
M = mass of earth = 5.98 x 10²⁴ kg
R = radius of earth = 6.37 x 10⁶ m
h = height above the earth's surface = 1400 km = 1.4 x 10⁶ m
r₁ = initial distance of the person from the center of earth when on surface = R = 6.37 x 10⁶ m
r₂ = final distance of the person from the center of earth when at some height = R + h = 6.37 x 10⁶ + 1.4 x 10⁶ = 7.77 x 10⁶ m
F₁ = Gravitational force of earth on the person when at surface
Gravitational force of earth on the person when at surface is given as
F₁ = G M m/r₁² eq-1
F₂ = Gravitational force of earth on the person when at some height
Gravitational force of earth on the person when at some height is given as
F₂ = G M m/r₂² eq-2
dividing eq-1 by eq-2
F₁ /F₂ = (G M m/r₁² )/(G M m/r₂²)
F₁ /F₂ = r₂²/r₁²
inserting the values
F₁ /F₂ = (7.77 x 10⁶)²/(6.37 x 10⁶)²
F₁ /F₂ = 1.49
