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All objects are attracted to the earth, and the direction of that force is towards the centre of the earth. The force exerted by the earth on an object is called the gravitational force (F). The magnitude of this force is what we know as the weight of the object. The acceleration a of an object shall now be replaced by g,…

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Acceleration Due to Gravity

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Jetzt kostenlos anmeldenAll objects are attracted to the earth, and the direction of that force is towards the centre of the earth. The force exerted by the earth on an object is called the **gravitational force** (F).

The magnitude of this force is what we know as the **weight **of the object. The acceleration a of an object shall now be replaced by g, which denotes **acceleration due to gravity**.

By **Newton’s second law of motion**, we know that:

\[F = m \cdot a \]

Here, a can be replaced by g, which gives us:

\[F = m \cdot g\]

This is the weight of the object under the influence of the gravity of the earth (often denoted by W). The unit of weight is the same as the force, which is N** **(called Newton, in honour of Sir Isaac Newton)** **or kg ⋅ m/s. Because it depends on g, the weight of any object depends on its geographical location.

For instance, even though the difference will be relatively small, the weight of an object with a certain mass will be more at sea level compared to its weight at the top of a mountain.

F is a vector quantity, as it has both magnitude and direction.

For a symmetrical object, the gravitational force** **acts towards the centre of the object. The value of g is almost constant near the surface of the earth, but as we move far from the surface of the earth, the strength of gravity decreases as the height increases.

The **acceleration **produced in any freely-falling body due to the **force of gravity **of another object, such as a planet,** **is known as **acceleration due to gravity**.

Based on experimental data, it has been observed that the **acceleration due to gravity **is inversely proportional to the square of the distance of the object from the centre of mass of the larger object.

\[g \propto \frac{1}{r^2}\]

Here, r is the distance of the object from the centre of the earth. The acceleration due to gravity is not only proportional inversely to r^2 but also directly proportional to the mass of the body attracted to, in this case, the earth.

For instance, the **acceleration due to gravity** **on the earth** is different from the **acceleration due to gravity on the moon**. Thus, we have another proportionality, as follows:

\[g \propto M\]

We assume that the mass of the object is significantly less with respect to the mass of the planet or body to which it is attracted. Algebraically, this is written as:

\[m << M\]

Here, *m = mass of the object *and *M = mass of the bigger object or planet*.

Combining both these proportionalities, we get:

\[g \propto \frac{M}{r^2}\]

To eliminate the proportionality and get equality, a\[g = \frac{GM}{r^2}\]

Based on experimental data, the value of G for the earth has been found to be G = 6.674⋅10^{-11 }Nm^{2} kg^{-2}.

Suppose the object is not on the surface of the earth but at a height h from the surface. In that case, its distance from the **centre of mass **of the earth will now be:

\[r = R + h\]

Here, R is the radius of the earth. Substituting for r in the earlier equation, we now get:

\[g = \frac{MG}{(R + h)^2}\]

(&)

** **

Hence, we can see that as h increases, the strength of gravity decreases.

The **acceleration due to gravity** does not follow the quadratic relationship when the object is below the surface of the earth. In fact, acceleration and distance are linearly dependent on each other for r < R (below the surface of the earth).

If an object is at r distance from the centre of the earth, the mass of the earth responsible for the **acceleration due to gravity **at that point will be:

\[m = \frac{Mr^3}{R^3}\]

This can be easily deduced using the formula for the volume of a sphere.

We have assumed the Earth to be a sphere, but in reality, the radius of the earth is at its minimum at the poles and at its maximum at the equator. The difference is quite small, and so we assume the earth to be a sphere for simplified calculations. The **acceleration due to gravity **follows the proportionality explained earlier:

\[g \propto \frac{m}{r^2}\]

Substituting for m, we get:

\[g = \frac{GMr}{R^3} g \propto r\]

We can now see that as G, M, and R are constants for a given object or planet, the acceleration linearly depends on r. Hence, we see that as r approaches R, the acceleration due to gravity increases according to the above linear relation, after which it decreases according to &**, **which we derived earlier. In practice, most of the real-world problems include the object being outside the surface of the earth.

The **acceleration due to gravity **has a linear relation with

This can be geometrically seen with the help of the graph above. As r increases, g reaches its maximum value when *r=R=radius of the earth*, and as we move away from the surface of the earth, the strength of g decreases according to the relation:

\[g \propto \frac{1}{r^2}\]

The equation describes a parabola, which is quite intuitive, given the definition we saw earlier.

We also note that the value of **acceleration due to gravity** is 0 at **the centre of the earth **and almost 0 **when**** far away from the surface of the earth. **To demonstrate the application of this concept, consider the following example.

The International Space Station, operating at an altitude of 35⋅10^{4} metres from the surface of the earth,** **plans to construct an object whose weight is 4.22⋅10^{6} N on the surface of the earth. What will be the weight of the same object once it arrives in the orbit of the Earth?

Note that g=9.81 ms^{-2}**, **the **radius of the earth, **R=6.37⋅10^{6 }m* , *and the

Apply the relevant equation, substitute the values provided, and solve for the unknown value. Sometimes, one equation is not enough, in which case solve for two equations, as the given data may not be enough to be directly substituted.

\[F = m \cdot g\]

\[g = \frac{MG}{r^2}\]

On the surface of the earth, we know that:

\[F = m \cdot g\]

\[\therefore m = \frac{F}{G}\]

\[m = \frac{4.22 \cdot 10^6 N}{9.81 m s^{-2}} m = 4.30 \cdot 10^5 kg\]

Now that we have determined the mass of the object, we need to use the formula of **acceleration due to gravity **to determine g** **at the orbital location:

**\[g = \frac{MG}{r^2}\]**

Now, we substitute the values, which gives us:

\[g = \frac{(5.97 \cdot 10^{24} kg) \cdot (6.674 \cdot 10^{-11} Nm^2 kg^{-2})}{(6.37 \cdot 10^6 m + 35 \cdot 10^4 m)^2}\]

And thus we have determined the** acceleration due to gravity **at the orbital location.

It should be noted that r is the distance from the centre of the earth, which requires our equation to be modified as follows:

*r = radius of the earth + distance of the orbit from the surface = R + h*

Now, we insert our calculated values for g and m in the initial formula for **weight**:

\[F = mg\]

\[F = (4.31 \cdot 10^5 kg) \cdot 8.82 ms^{-2} \qquad F = 3.80 \cdot 10^6 N\]

We now also know the **weight **of the object at the orbital location.

Don’t forget to specify the units of the quantity you are calculating, and always convert the data provided into similar units (preferably SI units).

- The direction of
**acceleration due to gravity**is always towards the centre of mass of the bigger object. **Acceleration due to gravity**is independent of the mass of the object itself and is only a function of its distance from the centre of mass of the bigger object.- The strength of gravity is maximum at the surface of the bigger object.
- The
**acceleration due to gravity**gradually decreases as we move far from the surface of the earth (or any object in general).

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