prove F = mg. how ? why?
Answers
Answer:
To answer your question, let's just place the formula F = G * m1 m2 / R^2 into a Taylor series expansion (as distance from the center of he earth is changed) and compare it with the approximation F = mg.
So, at Earth's surface
F = G m1 m2 / R^2
But if we to go just a little higher (say a height of h) than Earth's surface, we would have
F = G m1 m2 / (R+h)2
F = G m1 m2 / (R2 (1+h/R)2)
F = G m1 m2 / R2 * (1+h/R)-2
Now, take note that since F = mg is an approximation for the above formula at h << R, we can take a Taylor series expansion of (1+h/R)-2.
This expansion would be 1 - 2h/R + 3(h/R)2 - 4(h/R)3 +-... and so on.
Since h/R is close to 0 when h << R, we can just take the first term "1" of the Taylor series expansion to approximate F = G m1 m2 / R^2. From this expansion, we can also see that unless h is big enough to cause the next few terms "-2h/R" and "3(h/R)2" to be significant, F = mg is a pretty good approximation for the gravitational force at heights close to the Earth's radius.
I'd also like to show that the two formulas are numerically identical.
Formula 1: F = m * (g)
where F is the amount of force felt by an object of mass m, and g is the acceleration due to Earth's gravity near its surface.
Formula 2: F = m1 * (G * m2 / r2)
where F is the amount of force felt by an object of mass m1 due to an object with mass m2 at a distance of r apart from each other. G is the gravitational constant.
So, the numerical value of g is commonly used as 10 ms-2, or more accurately 9.81 ms-2.
G is gravitational constant, and its value is 6.67300 * 10-11 m3kg-1s-2.
m2 is actually the mass of the Earth, and its value is 5.9742 * 1024 kg.
r is actually the radius of the Earth, and its value is 6.3781 * 106 m.
Now, if you calculate the value of the term (G*m2/r^2), we get a value that is 9.7998 ms-2-- which is extremely close to the value of g (9.8ms-2)!
Hope that clears things up!
Explanation:
MARK ME BRAINLIEST
It is not. Newton’s second law states the force on an accelerating body is equal to the mass of the body times the acceleration. In other answers posted here you had that the force is the time derivative of the body’s momentum.
The formula in the question is none of those. It simply states that the force due to gravitational accelleration operating on a point mass at the earth surface at sea level on an average point on the equator is the mass of the point object times g. g is derived by measurement (average) or can be computed from neuton’s force formula F=mGMeR2e where G is newton gravitational constant Me is the mass of the earth Re is the earth radius (average taken as the equatorial average radius)