The force of gravity on earth due to the sun is 3.6 x 10^22N. What is the magnitude of the gravitational field due to the sun at earths location?

Question

The force of gravity on earth due to the sun is 3.6 x 10^22N. What is the magnitude of the gravitational field due to the sun at earths location?

What equation do I use for this, and how do I solve it?

Please help me, it would mean a lot.

Answer 1

We can look at the interaction of a planet in a gravitational interaction with another body, like a star like the Sun, a couple of different ways.

First, you can directly use Newton's Law of Universal Gravitation and calculate the force between the Earth and Sun. Let Me = 5.98 X 10²⁴ kg be the mass of the earth and Ms = 1.99 X 10³⁰ kg be the mass of the Sun. The center-to-center distance between the Earth and Sun is R = 1.5 X 10¹¹m. Newton's Law of Universal Gravitation is

F = GMeMs / R², where G = 6.67 X 10^-11 Nm² / kg² is Newton's Universal Gravitational constant,

F = (6.67 X 10^-11 Nm² / kg²)(5.98 X 10²⁴ kg)(1.99 X 10³⁰ kg)/(1.5 X 10¹¹m)² = 3.5 X 10²² N

This is how the number that you were given was calculated.

Another way to look at this problem is to see the Earth as a test mass in the Sun's gravitational field. Let the gravitational field of the Sun be given by a function A(R) = GMs / R², and any planet of mass M at a given distance R acts as a test mass in the Sun's gravitational field. For a given test mass M at a radial distance R from the Sun, the force would then be

F = M*A(R)

So, to get the value of the gravitational field at the location of the Earth, you would take the force value that you were given and divide it by the mass of the Earth:

A(R) = F/Me = 3.6 X 10²² N / 5.98 X 10²⁴ kg = 6.02 X 10^-3 N/kg