Near-Earth Gravitation

If we know the rough radius of a planet, as well as the approximate acceleration of gravity on an object close to its surface, we can apply Newton's Second Law to find the planet's mass:

Finding the Mass of Earth

Recall Newton's Second Law:

F_{net} = ma

Also recall the formula for gravitational force:

F_{ab} = G \frac{m_a m_b}{r_{ab}^2}

If we say object $a$ is some object near the surface and object $b$ is the Earth, we get:

F_{g} = G \frac{m_a m_E}{r_{E}^2}

Assuming gravity is the only force acting on the object, its net force $ma$ is equal to $mg$ :

m_ag = G \frac{m_a m_E}{r_{E}^2}

Notice $m_a$ cancels out on both sides, leaving us able to evaluate for gravity ( $g$ ):

g = G \frac{m_E}{r_{E}^2}

(1)

We know gravity is roughly $9.8 \text{ m/s}^2$ , the radius of Earth is approximately $6.37 \cdot 10^6 \text{ m}$ , and the gravitational constant $G$ is roughly $6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}$ , plugging these in we get:

9.8 \text{ m/s}^2 = \left(6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}\right) \frac{m_E}{(6.37 \cdot 10^6 \text{ m})^2}

Rearranging and solving for $m_E$ :

m_E = \frac{(9.8 \text{ m/s}^2)(6.37 \cdot 10^6 \text{ m})^2}{6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}} \approx 5.96 \cdot 10^{24} \text{ kg}

Example 1

Calculate the radius of Mars, given the planet's mass is approximately $6.417 \cdot 10^{23} \text{ kg}$ , and the acceleration caused by gravity on the surface is approximately $3.721 \text{ m/s}^2$ .

Solution Click to unblur

To find the radius of the planet, first rearrange Equation 1, solving for the planet's radius:

r_{M} = \sqrt{G \frac{m_{M}}{g}}

Plug in our known values for $m_M$ , $G$ , and $g$ , then solve:

r_{M} = \sqrt{\left( 6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2} \right) \frac{6.417 \cdot 10^{23} \text{ kg}}{3.721 \text{ m/s}^2}} \approx 3.39 \cdot 10^6 \text{ m}

Example 2

Calculate the acceleration induced by gravity on the International Space Station, which is roughly $400 \text{ km}$ above the surface of Earth.

Solution Click to unblur

The mass and radius of Earth are $5.97 \cdot 10^{24} \text{ kg}$ and $6.37 \cdot 10^6 \text{ m}$ , respectively.

To find the total distance from Earth's center, we must add the radius of Earth and the orbit altitude:

r = r_E + h_{\text{ISS}} = 6.37 \cdot 10^6 \text{ m} + 4 \cdot 10^5 \text{ m} = 6.77 \cdot 10^6 \text{ m}

Then plugging our values into Equation 1:

g = G \frac{m_E}{r^2} = \left( 6.67 \cdot 10^{-11} \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2} \right) \frac{5.97 \cdot 10^{24} \text{ kg}}{(6.77 \cdot 10^6 \text{ m})^2} \approx 8.68 \text{ m/s}^2

Apparent Weight at the Equator

Because Earth rotates, an object standing on the equator moves in a circle of radius $R_E$ and experiences a centripetal acceleration $a_c$ directed toward Earth's center:

a_c = \omega^2 R_E

Setting up Newton's Second Law for an object on a scale at the equator (where true gravity $F_g = mg$ points inward and the normal/scale force $F_N$ points outward):

F_{net} = mg - F_N = m a_c

Solving for your apparent weight (the normal force registered by the scale):

F_N = m(g - \omega^2 R_E)

Given Earth's angular velocity $\omega \approx 7.27 \cdot 10^{-5} \text{ rad/s}$ , this centripetal term reduces apparent acceleration by roughly $0.034 \text{ m/s}^2$ at the equator.