Respuesta :
Let
M=mass of earth
Object of mass m is at a distance r from center of earth, where r is greater than radius R of earth.
G=Universal gravitational constant
F=Gravitational force on the object
F = GMm/r^2
But object's acceleration is 'g*'
then
F=mg*
mg* = GMm/r^2
The acceleration due to gravity = g* = GM/r^2
At surface of earth = g =GM/R^2
g*/g =(R/r)^2
Distance from center of earth of a point one Earth radius above Earth's surface=r =2R
r = 2R
g*/g =(R/2R)^2
g*/g =(1/2)^2
g*/g =1/4
Weight at a point one Earth radius above Earth's
surface=mg*.....(a)
Weight at Earth's
surface=mg=100 N....(b)
Divide equation (a) by (b)
Weight at a point one Earth radius above Earth's
surface/weight on surface = mg*/mg =g*/g=1/4
Weight at a point one Earth radius above Earth's
surface=(1/4)*100=25 N
Weight at a point one Earth radius above Earth's
surface=25 N
M=mass of earth
Object of mass m is at a distance r from center of earth, where r is greater than radius R of earth.
G=Universal gravitational constant
F=Gravitational force on the object
F = GMm/r^2
But object's acceleration is 'g*'
then
F=mg*
mg* = GMm/r^2
The acceleration due to gravity = g* = GM/r^2
At surface of earth = g =GM/R^2
g*/g =(R/r)^2
Distance from center of earth of a point one Earth radius above Earth's surface=r =2R
r = 2R
g*/g =(R/2R)^2
g*/g =(1/2)^2
g*/g =1/4
Weight at a point one Earth radius above Earth's
surface=mg*.....(a)
Weight at Earth's
surface=mg=100 N....(b)
Divide equation (a) by (b)
Weight at a point one Earth radius above Earth's
surface/weight on surface = mg*/mg =g*/g=1/4
Weight at a point one Earth radius above Earth's
surface=(1/4)*100=25 N
Weight at a point one Earth radius above Earth's
surface=25 N
25 newtons.
Since the force of gravity varies inversely with the square of the distance. At distance R, the force is 100 Newtons. So we now have a distance 2R. So
(1) 100 = M/R^2
(2) X = M/(2R)^2
Solve equation (1) for M
100 = M/R^2
100R^2 = M
Substitute the value for M into equation (2) and solve
X = M/(2R)^2
X = (100R^2)/(2R)^2
X = (100R^2)/(4R^2)
X = (100)/(4)
X = 25
So the object at the higher altitude will weight 25 newtons.
Since the force of gravity varies inversely with the square of the distance. At distance R, the force is 100 Newtons. So we now have a distance 2R. So
(1) 100 = M/R^2
(2) X = M/(2R)^2
Solve equation (1) for M
100 = M/R^2
100R^2 = M
Substitute the value for M into equation (2) and solve
X = M/(2R)^2
X = (100R^2)/(2R)^2
X = (100R^2)/(4R^2)
X = (100)/(4)
X = 25
So the object at the higher altitude will weight 25 newtons.