If the mass of a planet is 3.80 1024 kg, and its radius is 2.20 106 m, what is the magnitude of the gravitational field, g, on the planet's surface?
F = m g = G m M /r^2
g = G M/r^2
where
G = 6.67 * 10^-11
M = 3.80 * 10^24
r = 2.20 * 10^6
A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the electron is a distance of 4.5 × 10−9 m from the gold nucleus.
To find the magnitude of the gravitational field, g, on the planet's surface, we can use the formula for gravitational field strength:
g = G * (m / r^2)
Where:
- g is the gravitational field strength
- G is the gravitational constant (approximately equal to 6.67430 × 10^-11 N m^2 / kg^2)
- m is the mass of the planet
- r is the radius of the planet
Let's substitute the values given:
m = 3.80 x 10^24 kg
r = 2.20 x 10^6 m
G = 6.67430 x 10^-11 N m^2 / kg^2
Plugging these values into the formula, we get:
g = (6.67430 x 10^-11 N m^2 / kg^2) * (3.80 x 10^24 kg) / (2.20 x 10^6 m)^2
First, let's square the radius:
r^2 = (2.20 x 10^6 m)^2 = 4.84 x 10^12 m^2
Now, let's substitute the value for m and r^2 back into the formula:
g = (6.67430 x 10^-11 N m^2 / kg^2) * (3.80 x 10^24 kg) / (4.84 x 10^12 m^2)
Now, let's calculate this using a calculator or a computer:
g ≈ 7.49 m/s^2
So, the magnitude of the gravitational field, g, on the planet's surface is approximately 7.49 m/s^2.