Acceleration Due to Gravity: Videos & Practice Problems

The acceleration due to gravity, denoted as g, can be calculated using Newton's law of gravitation. The formula is

g = G × M / r^2 for any distance, and g = G × M / R^2 at the surface. The weight of an object is the gravitational force, expressed as W = m × g. As distance from the planet increases, g decreases, illustrating the inverse-square law of gravitation.

Acceleration Due to Gravity

Understanding gravitational forces is essential in physics, particularly when calculating the acceleration due to gravity, denoted as g or sometimes g_surface. This acceleration can be derived from Newton's law of universal gravitation, which states that the gravitational force F between two masses is given by the formula:

$$ F = \frac{G m_1 m_2}{r^2} $$

where G is the gravitational constant, m₁ and m₂ are the masses of the objects, and r is the distance between their centers. When considering the acceleration due to gravity at a distance r from a planet's center, the formula simplifies to:

$$ g = \frac{G M}{r^2} $$

Here, M is the mass of the planet, and r is the distance from the center of the planet to the object. When an object is on the surface of the planet, r is equal to the planet's radius R, leading to the surface gravity equation:

$$ g_{surface} = \frac{G M}{R^2} $$

It is important to note that the acceleration due to gravity varies with distance from the planet. As the distance r increases, the value of g decreases, indicating that gravitational force weakens with distance. This relationship is crucial for understanding how weight changes with altitude.

Weight, defined as the force of gravity acting on an object, can be expressed as:

$$ W = mg $$

where m is the mass of the object and g is the acceleration due to gravity at that location. On the surface of the Earth, this becomes:

$$ W = mg_{surface} $$

To illustrate these concepts, consider calculating the acceleration due to gravity at a height, such as on Mount Everest. The height above the Earth's surface can be incorporated into the formula:

$$ g = \frac{G M}{(R + h)^2} $$

where h is the height above the surface. For Mount Everest, with a height of approximately 8,850 meters, the calculation yields a value of approximately 9.79 m/s², which is slightly less than the standard surface gravity of 9.81 m/s². This small difference highlights that even at significant altitudes, the effect on gravitational acceleration is minimal due to the vast radius of the Earth.

In summary, the acceleration due to gravity is a fundamental concept that varies with distance from a planet, and understanding its derivation and application is crucial for solving various physics problems related to gravitational forces.

You stand on the surface of a mysterious planet with a mass of 6 ×10²⁴ kg and measure the surface gravity to be 7 m/s². What must the radius of the planet be?

• A. 5.72×10¹³m
• B. 4.43×10¹⁴m
• C. 8.17×10¹²m
• D. 7.56×10⁶m

Answer: (D)

Find mass of planet in free fall

In this problem, we analyze the scenario of an astronaut dropping a rock from rest on the surface of an unknown planet. The goal is to determine the mass of the planet, denoted as \( M \). To achieve this, we can utilize the relationship between the acceleration due to gravity at the surface of the planet and its mass.

We start with the formula for gravitational acceleration at the surface, given by:

\[ g_{\text{surface}} = \frac{GM}{r^2} \]

Here, \( G \) is the gravitational constant, \( r \) is the radius of the planet, and \( g_{\text{surface}} \) is the acceleration due to gravity at the surface. To isolate \( M \), we rearrange the equation:

\[ M = \frac{g_{\text{surface}} \cdot r^2}{G} \]

Next, we need to determine \( g_{\text{surface}} \). The rock falls a distance of 1.5 meters in 0.6 seconds. This situation can be analyzed using kinematic equations. Since the rock is dropped from rest, its initial velocity \( v_0 \) is 0. The relevant kinematic equation is:

\[ \Delta y = v_0 t + \frac{1}{2} g t^2 \]

Substituting the known values, we have:

\[ 1.5 = 0 + \frac{1}{2} g (0.6)^2 \]

Solving for \( g \), we rearrange the equation:

\[ g = \frac{2 \cdot 1.5}{(0.6)^2} \]

Calculating this gives:

\[ g \approx 8.33 \, \text{m/s}^2 \]

Now that we have \( g_{\text{surface}} \), we can substitute it back into the mass equation. Given that the radius \( r \) of the planet is \( 4 \times 10^6 \) meters and \( G \) is approximately \( 6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \), we can find \( M \):

\[ M = \frac{8.33 \cdot (4 \times 10^6)^2}{6.67 \times 10^{-11}} \]

Calculating this yields:

\[ M \approx 2 \times 10^{24} \, \text{kg} \]

This result provides the mass of the unknown planet, demonstrating the application of kinematics and gravitational principles in solving real-world physics problems.

How far would you have to be above Earth's surface for g to be ½ of its surface value?

• A. 8.55×10⁸ m
• B. 2.64×10⁶ m
• C. 9.01×10⁶ m
• D. 8.13×10⁶ m

Answer: (B)

Here’s what students ask on this topic:

What is the formula for acceleration due to gravity?

The formula for acceleration due to gravity, denoted as g, can be derived from Newton's law of gravitation. For any distance from the center of a planet, it is given by:

$g=\frac{G\times M}{r^2}$

where G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet. At the surface of the planet, the formula simplifies to:

$g=\frac{G\times M}{R^2}$

where R is the radius of the planet.

How does the acceleration due to gravity change with distance from the Earth?

The acceleration due to gravity decreases as the distance from the Earth increases. This relationship follows the inverse-square law, which states that g is inversely proportional to the square of the distance r from the center of the Earth. Mathematically, it is expressed as:

$g=\frac{G\times M}{r^2}$

As r increases, the value of g decreases. This means that the farther you are from the Earth, the weaker the gravitational pull you experience.

What is the value of the acceleration due to gravity at the surface of the Earth?

The acceleration due to gravity at the surface of the Earth is approximately 9.81 meters per second squared (m/s²). This value is derived from the formula:

$g=\frac{G\times M}{R^2}$

where G is the gravitational constant (6.67 × 10^-11 N(m/kg)²), M is the mass of the Earth (5.97 × 10²⁴ kg), and R is the radius of the Earth (6.37 × 10⁶ m).

How do you calculate the weight of an object using the acceleration due to gravity?

The weight of an object can be calculated using the formula:

$W=m\times g$

where W is the weight, m is the mass of the object, and g is the acceleration due to gravity. At the surface of the Earth, g is approximately 9.81 m/s². For example, if an object has a mass of 10 kg, its weight would be:

$W=10\times 9.81=98.1 N$

where N stands for Newtons.

Why is the acceleration due to gravity considered a local constant at the Earth's surface?

The acceleration due to gravity at the Earth's surface is considered a local constant because it remains nearly the same regardless of where you are on the surface. This is due to the relatively small variations in altitude compared to the Earth's radius. The formula:

$g=\frac{G\times M}{R^2}$

shows that g depends on the Earth's mass M and radius R, both of which are constants. Therefore, g is approximately 9.81 m/s² everywhere on the Earth's surface, making it a local constant.