The value of g at the Earth's surface is about 9.8 m/s2.?

Question

What is the value of g at the distance of twice the Earth's radius?

Answer 1

Looking a Newtons law of Universal Gravitation,
F = G * M * m / r^2
and at Newton's second law of motion,
F = ma

we can arrive at a formula to find the acceleration due to some massive object's gravity,
a = G * M / r^2
where G is the Universal gravitational constant, M is the mass of the planet, and r is the radial distance one is observing the acceleration from the planet's center of mass.

In your question the only thing that is changing is the value of r, G and M are staying the same.
You can see that if you double the value of r, it gets squared in the formula (2^2 = 4) giving a r^2 value of 4 times what you originally had. Since the r^2 is in the denominator of the formula, as r grows, a shrinks.
By doubling r, the acceleration (g) is divided by 4.

You can look up the necessary values and plug everything into the formula to check your answer.

Answer 2

This is unfortunately impossible to answer specifically because it will vary a lot depending on all surrounding bodies, especially by our moon and Sun's positions at a specific time.

The force of attraction from the Earth will remain exactly the same according to Newtonian physics, but other forces may interact much more than within the atmosphere. As for the relativistic theory, the result would be fairly similar but considering additional factors will give more precise predictions.

Answer 3

its the same