Gravitational Force Near Earth

Main Idea

The gravitational force, as it acts near Earth's surface, is a simplified version of Newton's Law of Universal Gravitation. As a reminder, Newton's Law of Universal Gravitation states that any two bodies attract each other with a force that is dependent on the product of the two bodies' masses and is inversely dependent on the square of the distance between the two. Near Earth's surface, however, this can be simplified. The Earth is one of the bodies that we are interested in for determining this force, and because we know the mass of the Earth, the radius of the Earth, and the gravitational constant, we can simplify this force to an easier form. This is only applicable at distances near the Earth's surface, as we are using the radius of the Earth as the distance between the two objects. Generally this is acceptable because the distance between the object of interest and Earth's surface is [math]\displaystyle{ \lt \lt }[/math] the distance from the center of the Earth to Earth's surface. Thus this extra distance wouldn't affect the unsimplified calculation.

We use the radius of the Earth as well because when we are using the gravitational force, we simplify the bodies in question to point masses.

A Mathematical Model

The gravitational force in general is equal to:

[math]\displaystyle{ {F}_{grav}=G \frac{m_1 m_2}{r^2}\ }[/math]

where,

• F is the gravitational force between the masses;
• G is the gravitational constant, [math]\displaystyle{ 6.674×10^{−11} \frac{N m^2}{kg^2}\ }[/math];
• m₁ is the mass of the first object;
• m₂ is the mass of the second object;
• r is the distance between the centers of both masses.

Near Earth's surface, however, [math]\displaystyle{ \frac{GM_{Earth}}{R_{Earth}^2} }[/math] is equal to the gravitational constant near Earth's surface, g, which is equal to [math]\displaystyle{ 9.8 \frac{m}{s^2} }[/math]. This calculation is shown below:

[math]\displaystyle{ g = \frac{GM_{Earth}}{R_{Earth}^2}= \frac{(6.67*10^{-11} \frac{m^3}{kgs})(5.972*10^{24} kg)}{(6.378*10^{6} m)^2} = 9.8 \frac{m}{s^2} }[/math]

This then puts the force near Earth due to gravity into a much simpler form:

[math]\displaystyle{ |\vec{\mathbf{F}}_{grav}|= mg }[/math]

where,

• g is the near-Earth gravitational constant, as defined above to be [math]\displaystyle{ 9.8 \frac{m}{s^2} }[/math]
• m is the mass of the object whose behavior we are interested in

Examples

Simple

Suppose a 50 kg man is standing on the Earth. What force does gravity exert on the man?

Recall, [math]\displaystyle{ |\vec{\mathbf{F}}_{grav}|= mg }[/math]. Here, m is equal to 50 kg, and g is equal to [math]\displaystyle{ 9.8 \frac{m}{s^2} }[/math]. This means that:

[math]\displaystyle{ |\vec{\mathbf{F}}_{grav}|= mg = (50 kg)(9.8 \frac{m}{s^2}) = 490 N }[/math]

Hard

Suppose a 60 kg person is standing in an elevator that is accelerating upwards at [math]\displaystyle{ 1 \frac{m}{s^2} }[/math]. If they are standing on a scale, what does that scale read while the elevator accelerates upwards? A picture illustrating the situation is attached.

Here, the net acceleration is the same as that of the elevator, which is [math]\displaystyle{ 1 \frac{m}{s^2} }[/math]. The mass is equal to 60 kg. In order to determine the weight that the scale reads, we must look at the normal force exerted by the scale.

The above means that the net force is in the upwards direction.

[math]\displaystyle{ |\vec{\mathbf{F}}_{net}|=F_{N}-mg=ma }[/math]

[math]\displaystyle{ F_{N}=m(a+g) }[/math]

[math]\displaystyle{ F_{N}=(60 kg)(1 \frac{m}{s^2} +9.8 \frac{m}{s^2}) = 648 N }[/math]

Connectedness

The force of gravity near Earth's surface is relevant to so many problems. Whether throwing a ball, and watching it fall down, or watching two objects of drastically different weights drop to Earth's surface at the same acceleration, gravity is a unifying force. I am extremely interested in Space, and astrophysical events. Understanding how gravity acts near massive bodies (and the Earth with regards to asteroids and their acceleration) is important for understanding why bodies move the way they do.

I am a Physics major, and thus the force of gravity near Earth's surface connects to many of my classes. Furthermore, for a lot of engineering type majors, the force of gravity on objects that are being created (bridges, buildings, reactors), must be taken into consideration when deciding the type of material that object is created with and how much support must be provided to that building so that it doesn't collapse. Even when creating rockets, one must taken into account the force of gravity when deciding at what velocity that rocket must be shot off with in order to escape the Earth's gravitational pull.

With regards to industrial applications, as previously stated, the force of gravity must always be taken into account when building anything new, be it buildings or rockets. Ignoring the near-Earth gravitational force is a recipe for disaster.

History

Prior to relatively modern times, gravity as a force was a totally foreign concept. During the 17th century, Galileo had first realized that all objects fall with the same acceleration towards Earth, a hallmark of how gravity acts. It was in 1687 when Newton published his Principia that he hypothesized the inverse square law as it related to gravity. These calculations were incomplete, however, and remained as such until Cavendish performed his experiment and discovered the value of G, the gravitational constant.