The force due to gravity on a man is proportional to his mass and inversely proportional to the square of the distance from the centre of the earth.
If a man weighs 90kg and the radius of the earth is 6.4 x 10^6 metres, the force acting on him is 8 x 10^-4 Newton.
What is the force acting on a man weighing 80kg when he is in a satellite 1 million metres above the surface of the earth???
2007-08-04 14:54:01 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let F be the force
Let k be a constant
The force of gravity is proportional to his mass so it is related to him in the k*m manner
Gravity is inversely related to the square of the distance so it's related by the factor 1/r^2
F = km/r^2
8 x 10 ^-4 N = k * (90kg)/(6.4*10^6 m ) ^2
Solve for k.
Plug in the new values
F= k * (80kg)/(1,000,000)^2
Plug in what you got for k earlier and you should get the answer.
2007-08-04 15:29:48 · answer #1 · answered by Vu 3 · 0⤊ 0⤋
f=k m/d^2 = 8*10^-4 newton
k = (8*10^-4) (4.096*10^13/90) = 3.28* 10^10
F = 3.28*10^10 *80/7.4*10^6 = 3.28*10^10/5.48*10^ 13
= .598*10^-3 = 5.98*10^-4 newtons
2007-08-04 22:20:32 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋