### Learning Objectives

By the end of this section, you will be able to:

- Explain the connection between the constants $G$ and $g$
- Determine the mass of an astronomical body from free-fall acceleration at its surface
- Describe how the value of $g$ varies due to location and Earth’s rotation

In this section, we observe how Newton’s law of gravitation applies at the surface of a planet and how it connects with what we learned earlier about free fall. We also examine the gravitational effects within spherical bodies.

### Weight

Recall that the acceleration of a free-falling object near Earth’s surface is approximately $g=9.80\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2}$. The force causing this acceleration is called the weight of the object, and from Newton’s second law, it has the value *mg*. This weight is present regardless of whether the object is in free fall. We now know that this force is the gravitational force between the object and Earth. If we substitute *mg* for the magnitude of ${\overrightarrow{F}}_{12}$ in Newton’s law of universal gravitation, *m* for ${m}_{1}$, and ${M}_{\text{E}}$ for ${m}_{2}$, we obtain the scalar equation

$$mg=G\phantom{\rule{0ex}{0ex}}\frac{m{M}_{\text{E}}}{{r}^{2}}$$

where *r* is the distance between the centers of mass of the object and Earth. The average radius of Earth is about 6370 km. Hence, for objects within a few kilometers of Earth’s surface, we can take $r={R}_{\text{E}}$ (Figure 13.7). The mass *m* of the object cancels, leaving

$$g=G\frac{{M}_{\text{E}}}{{r}^{2}}.$$

13.2

This explains why all masses free fall with the same acceleration. We have ignored the fact that Earth also accelerates toward the falling object, but that is acceptable as long as the mass of Earth is much larger than that of the object.

Figure 13.7 We can take the distance between the centers of mass of Earth and an object on its surface to be the radius of Earth, provided that its size is much less than the radius of Earth.

### Example 13.3

#### Masses of Earth and Moon

Have you ever wondered how we know the mass of Earth? We certainly can’t place it on a scale. The values of *g* and the radius of Earth were measured with reasonable accuracy centuries ago.

- Use the standard values of
*g*, ${R}_{\text{E}}$, and Equation 13.2 to find the mass of Earth. - Estimate the value of
*g*on the Moon. Use the fact that the Moon has a radius of about 1700 km (a value of this accuracy was determined many centuries ago) and assume it has the same average density as Earth, $5500\phantom{\rule{0ex}{0ex}}{\text{kg/m}}^{3}$.

#### Strategy

With the known values of *g* and ${R}_{\text{E}}$, we can use Equation 13.2 to find ${M}_{\text{E}}$. For the Moon, we use the assumption of equal average density to determine the mass from a ratio of the volumes of Earth and the Moon.

#### Solution

- Rearranging Equation 13.2, we have
$${M}_{\text{E}}=\frac{g{R}_{\text{E}}^{2}}{G}=\frac{9.80\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2}{(6.37\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{6}\phantom{\rule{0ex}{0ex}}\text{m})}^{2}}{6.67\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{-11}}\phantom{\rule{0ex}{0ex}}\text{N}\xb7{\text{m}}^{2}{\text{/kg}}^{2}}=5.95\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{24}\phantom{\rule{0ex}{0ex}}\text{kg.}$$

- The volume of a sphere is proportional to the radius cubed, so a simple ratio gives us
$$\frac{{M}_{\text{M}}}{{M}_{\text{E}}}=\frac{{R}_{\text{M}}^{3}}{{R}_{\text{E}}^{3}}\to {M}_{\text{M}}=\left(\frac{{(1.7\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{6}\phantom{\rule{0ex}{0ex}}\text{m})}^{3}}{{(6.37\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{6}\phantom{\rule{0ex}{0ex}}\text{m})}^{3}}\right)(5.95\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{24}\phantom{\rule{0ex}{0ex}}\text{kg})=1.1\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{23}\phantom{\rule{0ex}{0ex}}\text{kg.}$$

We now use Equation 13.2.$${g}_{\text{M}}=G\phantom{\rule{0ex}{0ex}}\frac{{M}_{\text{M}}}{{r}_{\text{M}}^{2}}=(6.67\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{-11}}\phantom{\rule{0ex}{0ex}}\text{N}\xb7{\text{m}}^{2}{\text{/kg}}^{2})\frac{(1.1\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{23}\phantom{\rule{0ex}{0ex}}\text{kg})}{{(1.7\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{6}\phantom{\rule{0ex}{0ex}}\text{m})}^{2}}={2.5}^{}{\text{m/s}}^{2}$$

#### Significance

As soon as Cavendish determined the value of *G* in 1798, the mass of Earth could be calculated. (In fact, that was the ultimate purpose of Cavendish’s experiment in the first place.) The value we calculated for *g* of the Moon is incorrect. The average density of the Moon is actually only $3340\phantom{\rule{0ex}{0ex}}{\text{kg/m}}^{3}$ and $g=1.6\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2}$ at the surface. Newton attempted to measure the mass of the Moon by comparing the effect of the Sun on Earth’s ocean tides compared to that of the Moon. His value was a factor of two too small. The most accurate values for *g* and the mass of the Moon come from tracking the motion of spacecraft that have orbited the Moon. But the mass of the Moon can actually be determined accurately without going to the Moon. Earth and the Moon orbit about a common center of mass, and careful astronomical measurements can determine that location. The ratio of the Moon’s mass to Earth’s is the ratio of [the distance from the common center of mass to the Moon’s center] to [the distance from the common center of mass to Earth’s center].

Later in this chapter, we will see that the mass of other astronomical bodies also can be determined by the period of small satellites orbiting them. But until Cavendish determined the value of *G*, the masses of all these bodies were unknown.

### Example 13.4

#### Gravity above Earth’s Surface

What is the value of *g* 400 km above Earth’s surface, where the International Space Station is in orbit?

#### Strategy

Using the value of ${M}_{\text{E}}$ and noting the radius is $r={R}_{\text{E}}+400\phantom{\rule{0ex}{0ex}}\text{km}$, we use Equation 13.2 to find *g*.

From Equation 13.2 we have

$$g=G\frac{{M}_{\text{E}}}{{r}^{2}}=6.67\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{-11}}\phantom{\rule{0ex}{0ex}}\text{N}\xb7{\text{m}}^{2}{\text{/kg}}^{2}\frac{5.96\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{24}\phantom{\rule{0ex}{0ex}}\text{kg}}{{(6.37\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{6}+400\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{3}\phantom{\rule{0ex}{0ex}}\text{m})}^{2}}=8.67\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2}.$$

#### Significance

We often see video of astronauts in space stations, apparently weightless. But clearly, the force of gravity is acting on them. Comparing the value of *g* we just calculated to that on Earth $(9.80\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2})$, we see that the astronauts in the International Space Station still have 88% of their weight. They only appear to be weightless because they are in free fall. We will come back to this in Satellite Orbits and Energy.

### Check Your Understanding 13.2

How does your weight at the top of a tall building compare with that on the first floor? Do you think engineers need to take into account the change in the value of *g* when designing structural support for a very tall building?

### The Gravitational Field

Equation 13.2 is a scalar equation, giving the magnitude of the gravitational acceleration as a function of the distance from the center of the mass that causes the acceleration. But we could have retained the vector form for the force of gravity in Equation 13.1, and written the acceleration in vector form as

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

We identify the vector field represented by $\overrightarrow{g}$ as the gravitational field caused by mass $M$. We can picture the field as shown Figure 13.8. The lines are directed radially inward and are symmetrically distributed about the mass.

Figure 13.8 A three-dimensional representation of the gravitational field created by mass $M$. Note that the lines are uniformly distributed in all directions. (The box has been added only to aid in visualization.)

As is true for any vector field, the direction of $\overrightarrow{g}$ is parallel to the field lines at any point. The strength of $\overrightarrow{g}$ at any point is inversely proportional to the line spacing. Another way to state this is that the magnitude of the field in any region is proportional to the number of lines that pass through a unit surface area, effectively a density of lines. Since the lines are equally spaced in all directions, the number of lines per unit surface area at a distance *r* from the mass is the total number of lines divided by the surface area of a sphere of radius *r*, which is proportional to ${r}^{2}$. Hence, this picture perfectly represents the inverse square law, in addition to indicating the direction of the field. In the field picture, we say that a mass *m* interacts with the gravitational field of mass *M*. We will use the concept of fields to great advantage in the later chapters on electromagnetism.

### Apparent Weight: Accounting for Earth’s Rotation

As we saw in Applications of Newton’s Laws, objects moving at constant speed in a circle have a centripetal acceleration directed toward the center of the circle, which means that there must be a net force directed toward the center of that circle. Since all objects on the surface of Earth move through a circle every 24 hours, there must be a net centripetal force on each object directed toward the center of that circle.

Let’s first consider an object of mass *m* located at the equator, suspended from a scale (Figure 13.9). The scale exerts an upward force ${\overrightarrow{F}}_{\text{s}}$ away from Earth’s center. This is the reading on the scale, and hence it is the apparent weight of the object. The weight (*mg*) points toward Earth’s center. If Earth were not rotating, the acceleration would be zero and, consequently, the net force would be zero, resulting in ${F}_{\text{s}}=mg$. This would be the true reading of the weight.

Figure 13.9 For a person standing at the equator, the centripetal acceleration $({a}_{\text{c}})$ is in the same direction as the force of gravity. At latitude $\lambda $, the angle the between ${a}_{\text{c}}$ and the force of gravity is $\lambda $ and the magnitude of ${a}_{\text{c}}$ decreases with $\text{cos}\lambda $.

With rotation, the sum of these forces must provide the centripetal acceleration, ${a}_{\text{c}}$. Using Newton’s second law, we have

$$\sum F={F}_{\text{s}}}-mg=m{a}_{\text{c}}\phantom{\rule{0ex}{0ex}}\text{where}\phantom{\rule{0ex}{0ex}}{a}_{\text{c}}=-\frac{{v}^{2}}{r}.$$

13.3

Note that ${a}_{\text{c}}$ points in the same direction as the weight; hence, it is negative. The tangential speed *v* is the speed at the equator and *r* is ${R}_{\text{E}}$. We can calculate the speed simply by noting that objects on the equator travel the circumference of Earth in 24 hours. Instead, let’s use the alternative expression for ${a}_{\text{c}}$ from Motion in Two and Three Dimensions. Recall that the tangential speed is related to the angular speed $\left(\omega \right)$ by $v=r\omega $. Hence, we have ${a}_{c}=\text{\u2212}r{\omega}^{2}$. By rearranging Equation 13.3 and substituting $r={R}_{\text{E}}$, the apparent weight at the equator is

$${F}_{\text{s}}=m\left(g-{R}_{\text{E}}{\omega}^{2}\right).$$

The angular speed of Earth everywhere is

$$\omega =\frac{2\pi \phantom{\rule{0ex}{0ex}}\text{rad}}{24\phantom{\rule{0ex}{0ex}}\text{hr}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}3600\phantom{\rule{0ex}{0ex}}\text{s/hr}}=7.27\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{\mathrm{-5}}\phantom{\rule{0ex}{0ex}}\text{rad/s.}$$

Substituting for the values or ${R}_{\text{E}}$ and $\omega $, we have ${R}_{\text{E}}{\omega}^{2}=0.0337\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2}$. This is only 0.34% of the value of gravity, so it is clearly a small correction.

### Example 13.5

#### Zero Apparent Weight

How fast would Earth need to spin for those at the equator to have zero apparent weight? How long would the length of the day be?

#### Strategy

Using Equation 13.3, we can set the apparent weight (${F}_{\text{s}}$) to zero and determine the centripetal acceleration required. From that, we can find the speed at the equator. The length of day is the time required for one complete rotation.

#### Solution

From Equation 13.2, we have $\sum F={F}_{\text{s}}}-mg=m{a}_{\text{c}$, so setting ${F}_{\text{s}}=0$, we get $g={a}_{\text{c}}$. Using the expression for ${a}_{\text{c}}$, substituting for Earth’s radius and the standard value of gravity, we get

$$\begin{array}{}\\ \\ {a}_{\text{c}}=\frac{{v}^{2}}{r}=g\hfill \\ v=\sqrt{gr}=\sqrt{(9.80\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2})(6.37\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{6}\phantom{\rule{0ex}{0ex}}\text{m})}=7.91\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{3}\phantom{\rule{0ex}{0ex}}\text{m/s}.\hfill \end{array}$$

The period *T* is the time for one complete rotation. Therefore, the tangential speed is the circumference divided by *T*, so we have

$$\begin{array}{}\\ v=\frac{2\pi r}{T}\hfill \\ T=\frac{2\pi r}{v}=\frac{2\pi (6.37\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{6}\phantom{\rule{0ex}{0ex}}\text{m})}{7.91\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{3}\phantom{\rule{0ex}{0ex}}\text{m/s}}=5.06\phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}{10}^{3}\phantom{\rule{0ex}{0ex}}\text{s}.\hfill \end{array}$$

This is about 84 minutes.

#### Significance

We will see later in this chapter that this speed and length of day would also be the orbital speed and period of a satellite in orbit at Earth’s surface. While such an orbit would not be possible near Earth’s surface due to air resistance, it certainly is possible only a few hundred miles above Earth.

### Results Away from the Equator

At the poles, ${a}_{\text{c}}\to 0$ and ${F}_{\text{s}}=mg$, just as is the case without rotation. At any other latitude $\lambda $, the situation is more complicated. The centripetal acceleration is directed toward point *P* in the figure, and the radius becomes $r={R}_{\text{E}}\text{cos}\lambda $. The *vector* sum of the weight and ${\overrightarrow{F}}_{\text{s}}$ must point toward point *P*, hence ${\overrightarrow{F}}_{\text{s}}$ no longer points away from the center of Earth. (The difference is small and exaggerated in the figure.) A plumb bob will always point along this deviated direction. All buildings are built aligned along this deviated direction, not along a radius through the center of Earth. For the tallest buildings, this represents a deviation of a few feet at the top.

It is also worth noting that Earth is not a perfect sphere. The interior is partially liquid, and this enhances Earth bulging at the equator due to its rotation. The radius of Earth is about 30 km greater at the equator compared to the poles. It is left as an exercise to compare the strength of gravity at the poles to that at the equator using Equation 13.2. The difference is comparable to the difference due to rotation and is in the same direction. Apparently, you really can lose “weight” by moving to the tropics.

### Gravity Away from the Surface

Earlier we stated without proof that the law of gravitation applies to spherically symmetrical objects, where the mass of each body acts as if it were at the center of the body. Since Equation 13.2 is derived from Equation 13.1, it is also valid for symmetrical mass distributions, but both equations are valid only for values of $r\ge {R}_{\text{E}}$. As we saw in Example 13.4, at 400 km above Earth’s surface, where the International Space Station orbits, the value of *g* is $8.67\phantom{\rule{0ex}{0ex}}{\text{m/s}}^{2}$. (We will see later that this is also the centripetal acceleration of the ISS.)

For $r<{R}_{\text{E}}$, Equation 13.1 and Equation 13.2 are not valid. However, we can determine *g* for these cases using a principle that comes from Gauss’s law, which is a powerful mathematical tool that we study in more detail later in the course. A consequence of Gauss’s law, applied to gravitation, is that only the mass *within r* contributes to the gravitational force. Also, that mass, just as before, can be considered to be located at the center. The gravitational effect of the mass *outside r* has zero net effect.

Two very interesting special cases occur. For a spherical planet with constant density, the mass within *r* is the density times the volume within *r*. This mass can be considered located at the center. Replacing ${M}_{\text{E}}$ with only the mass within *r*, $M=\rho \phantom{\rule{0ex}{0ex}}\times \phantom{\rule{0ex}{0ex}}(\text{volume of a sphere})$, and ${R}_{\text{E}}$ with *r*, Equation 13.2 becomes

$$g=G\phantom{\rule{0ex}{0ex}}\frac{{M}_{\text{E}}}{{R}_{\text{E}}^{2}}=G\phantom{\rule{0ex}{0ex}}\frac{\rho \left(4\text{/}3\pi {r}^{3}\right)}{{r}^{2}}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}G\rho \pi r.$$

The value of *g*, and hence your weight, decreases linearly as you descend down a hole to the center of the spherical planet. At the center, you are weightless, as the mass of the planet pulls equally in all directions. Actually, Earth’s density is not constant, nor is Earth solid throughout. Figure 13.10 shows the profile of *g* if Earth had constant density and the more likely profile based upon estimates of density derived from seismic data.

Figure 13.10 For $r<{R}_{\text{E}}$, the value of *g* for the case of constant density is the straight green line. The blue line from the PREM (Preliminary Reference Earth Model) is probably closer to the actual profile for *g*.

The second interesting case concerns living on a spherical shell planet. This scenario has been proposed in many science fiction stories. Ignoring significant engineering issues, the shell could be constructed with a desired radius and total mass, such that *g* at the surface is the same as Earth’s. Can you guess what happens once you descend in an elevator to the inside of the shell, where there is no mass between you and the center? What benefits would this provide for traveling great distances from one point on the sphere to another? And finally, what effect would there be if the planet was spinning?

## FAQs

### What is the equation for gravity near the surface of the earth? ›

Newton's law of gravitation is: **F = GMm r2** where the Gravitational Constant G = 6.673 × 10−11Nm2kg−2 (kg−1m3s−2). gravitational force per unit mass = gravitational acceleration g. g is approximately 9.8m/s2 at the surface of the Earth.

**What is the magnitude of force of gravitation due to Earth on a ball of 1 kg mass lying on the ground? ›**

Hence, the gravitational force between the ball and the earth is **9.8 Newton**.

**How do you calculate g? ›**

The formula of gravity is indicated by **G = GM / R ^{2}**.

**Is gravity 9.8 meters per second squared at the earth's surface? ›**

A: Surface gravity is the gravity felt at the surface of a planet. Earth's surface grvaity is 9.8 meters per second squared (**32.2 feet per second squared**).

**What is the formula of acceleration and due to gravity on the earth surface? ›**

Based on Newton and Cavendish's research, scientists developed the acceleration due to gravity formula: **g = G M r 2** , where: G is the gravitational constant 6.67 × 10 − 11 N m 2 / k g 2 , M is the mass of the object, and r is the radius of the object.

**What is the magnitude of the gravitational force between the earth and 1? ›**

Hence, the magnitude of the gravitational force between the earth and a 1 kg object on its surface is **9.77 N**.

**What is the magnitude of gravitational force between earth and an object of 1 kg? ›**

This shows that Earth exerts a force of **9.8 N** on a body of mass 1 kg.

**What is the gravitational force of earth acting on a mass of 1 kg? ›**

Hence, **9.8N** is the force with which a body of mass 1 kg is attracted towards the earth. Q.

**What is g value in physics? ›**

The value of G is **(6.6743 ± 0.00015) × 10 ^{−}^{11} m^{3} kg^{−}^{1} s^{−}^{2}**. Related Topics: gravity physical constant Cavendish experiment gravitation. See all related content → It must be pointed out that G occupies a rather anomalous position among the other constants of physics.

**How much force is 1g? ›**

A G-force of 1g is equal to the conventional value of gravitational acceleration on Earth which amounts to **9.8 m/s ^{2} (meter per second squared)**. In other words, it is the force a person (driver, in our case) can experience when they're accelerating or braking.

### What if Earth had no gravity for 1 second? ›

Nothing much. If it is only for one second, then depending on where you are on Earth **you would start floating for one second going up somewhat** (but not very much). i.e. about 3cm. Some objects will probably break when the gravity is back on and they fall back to the surface.

**At what distance from the earth surface is gravity zero? ›**

If the earth were about 36,000 km in diameter with the same mass and length-of-day then the gravity at **the equator** would be zero. This is the altitude of geostationary orbits.

**What does 9.8 mean in physics? ›**

The numerical value for the **acceleration of gravity** is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude.

**What is the gravity at a depth? ›**

The formula for g at depth d

The formula **g2= g (1 – d/R)** expresses the variation of g with depth. Here, g2 denotes the acceleration due to gravity at a depth of d from the earth's surface, and R denotes the earth's radius.

**Does gravity change with altitude? ›**

**gravity increases with height**. gravity is significantly less on high mountains or tall buildings and increases as we lose height (which is why falling objects speed up) gravity is caused by the Earth spinning. gravity affects things while they are falling but stops when they reach the ground.

**What does the value of g on the earth surface depend on? ›**

Therefore, acceleration due to gravity (ggg) depends on **the mass of the planet, distance between centre of planet and object, shape of the planet, rotational parameters of the planet**.

**What is the gravitational force between Earth and an object of 2 kg mass? ›**

So Newton discovered this gravitational force and due to this everybody on the earth is acted upon by the force due to the earth and this is what we call as the weight of the body. Where 'G' is the gravitational constant. Hence 2kg mass will experience **19.6N** of force.

**What is the magnitude of the gravitational force exerted by a 15kg mass? ›**

a_{2} = Gm10.252 =6.673∗10−11∗150.252= **1.6*10−8 ms2**.

**How does the force of gravity between two bodies change when the distance between them is doubled? ›**

As Force is inversely proportional to square of distance between them, force will **reduce to one fourth the initial force** when distance between them is doubled.

**What is Newton's equation for gravity? ›**

In symbols, the magnitude of the attractive force F is equal to G (the gravitational constant, a number the size of which depends on the system of units used and which is a universal constant) multiplied by the product of the masses (m_{1} and m_{2}) and divided by the square of the distance R: **F = G(m _{1}m_{2})/R^{2}**.

### What is the gravitational force between two 1kg objects that are 1 m apart? ›

Hence, the force between two bodies each weighing 1 kg and separated 1 m apart will be. **67 × 10 - 11 N**.

**What is the magnitude of the gravitational force exerted by? ›**

The magnitude of the gravitational force will be **equal to the product of mass and acceleration due to the gravitational force**. The Universal Law of Gravitation says that – any two objects having mass will attract each other with a force which is directly proportional to the product of their masses.

**What is the magnitude of Earth's gravitational force on a 1 kg body at Earth's surface quizlet? ›**

What is the magnitude of earth's gravitational force on 1kg body at earth's surface? 6. The gravitational force is about 10 N, or more accurately, **9.8 N**. When G was first measured by Henry Cavnedish, newspapers of the time hailed his experiment as the "weighing earth experiment," Why?

**What is the force of gravity acting on a 1 kg mass quizlet? ›**

1 Kg=**9.8N**. 1 Kg~10N. Strength of gravity is 9.8N/Kg, that means- Each kg has a mass weighs 9.8 newtons.

**What is the weight of 1 kilogram mass on the earth g is equal to 9.8 m per second square? ›**

= 1 kg x 9.8m/s^{2}=**9.8 N**.

**Is g always 9.8 in physics? ›**

"9.81 meters per second squared" means that objects on Earth will accelerate (or go faster) 9.81 meters every second, if they are in free fall, due to the pull of gravity. **Throughout space, gravity actually is constant**.

**What unit is g in physics? ›**

Gravitational constant | |
---|---|

Common symbols | G |

SI unit | m^{3}⋅kg^{−}^{1}⋅s^{−}^{2} |

Dimension |

**What is the value of g in physics gravity? ›**

It is defined by standard as 9.80665 m/s^{2} (about 32.17405 ft/s^{2}). This value was established by the 3rd General Conference on Weights and Measures (1901, CR 70) and used to define the standard weight of an object as the product of its mass and this nominal acceleration.

**Is gravity a law or theory? ›**

Isaac Newton's 1687 description of gravity was considered **scientific law** until Einstein's General Theory of Relativity, published more than two centuries later. Newton had explained gravity as a force that instantaneously acts over a distance. The result is a pull between any two objects in the universe.

**Why does mass create gravity? ›**

According to theory, the reason mass is proportional to gravity is because **everything with mass emits tiny particles called gravitons**. These gravitons are responsible for gravitational attraction. The more mass, the more gravitons.

### What is 10g gravity? ›

Kirby often says, “At 10 G's, **it's hard to breathe and feels like a house is sitting on your chest**.” That's a ton of pressure, literally! When a person begins to lose consciousness due to positive G's, this is called G-lock.

**What does 3 gs feel like? ›**

For example, let's say that there is a boy that ways 100 pounds. That is his weight at one G. At two G's, the boy will feel like he weighs 200 pounds. At three G's, he will feel like he weighs 300 pounds.

**What does 9 gs feel like? ›**

Under 9G's, the world appears to shrink until it looks like you're viewing it through a toilet paper roll. Blood is being pulled out of your head towards your legs and arms, resulting in the **loss of peripheral vision**.

**What if gravity stopped for 1 minute? ›**

If the Earth's gravity did just suddenly disappear **we would no longer have a force keeping us on the ground**. The Earth would keep spinning, as it does, but we would no longer move with it; instead we would move in a straight line, upwards.

**What if gravity stopped for 5 seconds? ›**

As every object will be out of balance for those 5 seconds, the earth's atmosphere will start to disappear, its core will expand with the heat of the sun, the surface of the earth will begin to crack, tidal waves will soar high and other sudden changes will start coming to life.

**What would happen if gravity reversed? ›**

It would **push objects away from each other**. Watching this happen from out in space, you could see everything not bolted down to Earth–buildings, desks, homework, cats–start to lift off and drift into space. Then, you could see the surface of the earth start to fall away.

**At what height do we lose gravity? ›**

If you wanted to reach a point where Earth's gravity no longer has a hold on you, you'd have to fly out about **21 million kilometers**, or 13 million miles.

**What speed is needed to break Earth's gravity? ›**

The amount of velocity needed to escape our planets gravity is approx **11km per second**, which is known as escape velocity.

**Does gravity get stronger the deeper you go? ›**

As you go down below the Earth's surface, in a mine shaft for example, **the force of gravity lessens**. Weight and gravitational pull continue to decrease as you get closer to the centre of the Earth.

**Where is the lowest gravity on Earth? ›**

World Least Gravity- **Hiriwadunna, Sri Lanka**

Sri Lanka and south India has least gravity level on the earth. As you say this is almost the equatorial region and this area has minimal gravity as measured by the Sri Lankan government. The point on the Earth's surface with the lowest gravity is Hiriwadunna in Sri Lanka.

### How much g is on Earth? ›

Gravity is measured as how fast objects accelerate towards each other. The average gravitational pull of the Earth is **9.8 meters per second squared** (m/s2).

**Is gravity a force or acceleration? ›**

**Gravity is the force that pulls an object towards the center of the earth**. The value of the acceleration due to the gravity on earth is 9.8 m/s2. g = GM/r2 is the equation used to calculate acceleration due to gravity.

**What is the magnitude of the gravitational force between the earth and a 10kg object on its surface? ›**

Gravitational force on a body on the surface of the earth = mass x acceleration due to gravity = 10 kg x 9.8 ms^{-}^{2} = **98 N**.

**What is the magnitude of the gravitational force between the earth and a 2kg object on its surface? ›**

So Newton discovered this gravitational force and due to this everybody on the earth is acted upon by the force due to the earth and this is what we call as the weight of the body. Where 'G' is the gravitational constant. Hence 2kg mass will experience **19.6N** of force.

**What is the weight of 1 gram of object at the earth's surface? ›**

Since g = 9.80 m/s^{2} on Earth, the weight of a 1.00-kg object on Earth is **9.80 N**: w=mg=(1.00kg)(9.80m/s2)=9.80N.

**What is a 1 kg mass at the earth's surface? ›**

A 1.0kg mass weighs **9.8n** on Earth's surface, and the radius of the Earth is roughly 6.4x10 m.

**What is the magnitude of the earth's gravitational field strength on the earth's surface in N kg? ›**

The magnitude of the gravitational field at the surface of the earth is around **9.8 N kg ^{-}^{1}**.