Lesson Explainer: Surface Gravity Physics • 9th Grade

In this explainer, we will learn how to calculate the surface gravity of a planet or moon given its mass and its radius.

From Newton’s second law of motion, recall that if a force is acting on an object, that object will experience acceleration proportional to the magnitude of the net force. We can write this law as

⃑𝐹=π‘šβƒ‘π‘Ž,(1)

where ⃑𝐹 is the force acting on the object, π‘š is the object’s mass, and βƒ‘π‘Ž is the acceleration. Notice that ⃑𝐹 and βƒ‘π‘Ž are vectors, meaning that they have both magnitude and direction. This indicates that the acceleration experienced is in the same direction as the force.

Imagine two isolated objects in deep space, with no stars, planets, or anything else nearby, so that the only force acting on each of them is the gravitational force due to each other. Recall from Newton’s law of gravitation that the gravitational force, ⃑𝐹, is written as

⃑𝐹=πΊπ‘€π‘šβ€–β€–βƒ‘π‘Ÿβ€–β€–βƒ‘π‘Ÿ,(2)

where 𝐺 is the universal gravitational constant 𝐺=6.67Γ—10/β‹…ο…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs, 𝑀 and π‘š are the masses of the two objects, and βƒ‘π‘Ÿ is the distance between the centers of mass of the two objects. The notation β€–β€–βƒ‘π‘Ÿβ€–β€– indicates the magnitude of a vector; in this case, β€–β€–βƒ‘π‘Ÿβ€–β€– is the distance. To specify the direction, we have βƒ‘π‘Ÿ, which is the unit vector in the direction connecting the two objects’ centers of mass. This tells us that the force, ⃑𝐹, acts along the line connecting the two objects’ centers of mass.

Notice that the force experienced by both objects is the same and is proportional to the product of their masses.

If the gravitational force is the only force acting on the objects, then the force ⃑𝐹, in equation (2), is equal to the net force ⃑𝐹, in equation (1). We can therefore equate the right-hand sides of those equations so that π‘šβƒ‘π‘Ž=πΊπ‘€π‘šβ€–β€–βƒ‘π‘Ÿβ€–β€–βƒ‘π‘Ÿ.

Here, π‘š, the mass of the object whose acceleration we are considering, appears on both sides of this equation. This means we can divide both sides by π‘š, and we are left with

βƒ‘π‘Ž=πΊπ‘€β€–β€–βƒ‘π‘Ÿβ€–β€–βƒ‘π‘Ÿ.(3)

So the acceleration, βƒ‘π‘Ž, depends only on the mass of the other object, which we have called 𝑀, and the distance, β€–β€–βƒ‘π‘Ÿβ€–β€–, between their centers of mass. This means that the acceleration due to gravity of an object does not depend on the mass of that object, only on the mass of the other object whose gravitational force it is experiencing.

Note that here, too, both βƒ‘π‘Ž and βƒ‘π‘Ÿ are vectors, so they have both magnitude and direction. We measure βƒ‘π‘Ÿ from the centers of mass of each object, and the acceleration acts along that same line: each object experiences acceleration toward the center of mass of the other object.

In most cases, we are only interested in the magnitude of the acceleration. We can also write equation (3) in scalar form as

π‘Ž=πΊπ‘€π‘Ÿ,(4)

where we have indicated that acceleration, π‘Ž, and distance, π‘Ÿ, are scalars by writing them in regular font. The notation π‘Ÿ is equivalent to β€–β€–βƒ‘π‘Ÿβ€–β€–. When written like this, we are only calculating the magnitude of the acceleration, but the direction is still always toward the center of mass of the other object.

Let’s have a look at an example.

Example 1: Finding the Gravitational Force and Acceleration due to Gravity for Objects of Different Mass

Two objects, object A with a mass of 5 kg and object B with a mass of 100 kg, are near an even larger object with a mass of 10 kg. Object A and object B are at an equal distance away, 100 km, from the center of mass of the very large object.

  1. What is the magnitude of the gravitational force experienced by object A due to the very large object? Give your answer to two decimal places.
  2. What is the acceleration of object A toward the very large object? Give your answer to two decimal places.
  3. What is the magnitude of the gravitational force experienced by object B due to the very large object? Give your answer to two decimal places.
  4. What is the acceleration of object B toward the very large object? Give your answer to two decimal places.

Answer

In this example, we have a very large object who mass is significantly more than that of either of the two objects A and B. Although it is depicted as a rectangle, we are told that it is spherical, so it must be so large that we can see only a small portion of it.

To give an idea of the scales involved, the mass of the very large object is equivalent to a small moon, object B is approximately the mass of a person, and object A is about the mass of a large book. The very large object has so much more mass than objects A and B that we can neglect any gravitational forces that objects A and B may exert on each other. The dominant forces will be between each object and the very large object.

Part 1

The first part of the question asks for the magnitude of the gravitational force experienced by object A due to the very large object, so we need to recall equation (2) for gravitational force, 𝐹, which is 𝐹=πΊπ‘€π‘šπ‘Ÿ, where 𝐺=6.67Γ—10/β‹…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs is the universal gravitational constant, 𝑀=10kg is the mass of the very large object, π‘š=5kg is the mass of object A, and π‘Ÿ=100km is the distance of object A from the center of mass of the very large object.

In order for the force to be in newtons, we need to ensure we input each quantity using the correct units. The mass units in both 𝐺 and the masses 𝑀 and π‘š are in kilograms, which is correct, but we have the distance π‘Ÿ in kilometres rather than metres. To convert π‘Ÿ to the correct units, we need to recall that 1=1000kmm, so π‘Ÿ=100=100000kmm.

Now we can input these values into the equation for 𝐹, and we find that 𝐹=πΊπ‘€π‘šπ‘Ÿ=6.67Γ—10/β‹…Γ—10Γ—5(100000)=3.335.mkgskgkgmN

So, to two decimal places, the final answer is 𝐹=3.34.N

Part 2

The next part of the question asks for the acceleration of object A toward the very large object. To solve this, we need to refer to equation (4) for acceleration, π‘Ž, due to gravity: π‘Ž=πΊπ‘€π‘Ÿ, where 𝐺=6.67Γ—10/β‹…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs is the universal gravitational constant, 𝑀=10kg is the mass of the very large object, and π‘Ÿ=100000m. As above, we are using the value of π‘Ÿ in metres to ensure that the acceleration, π‘Ž, is in the desired unit of metres per second squared (m/s2).

Using these values in the equation for π‘Ž, we find π‘Ž=πΊπ‘€π‘Ÿ=6.67Γ—10/β‹…Γ—10(100000)=0.667/.mkgskgmms

Rounding to two decimal places, we have π‘Ž=0.67/.ms

Part 3

In the third part of the question, we need to calculate the magnitude of the gravitational force experienced by object B due to the very large object. We will approach this in the same way we did for object A, using the equation for gravitational force 𝐹=πΊπ‘€π‘šπ‘Ÿ, but this time, π‘š is the mass of object B, so we will use a value of 100 kg. Substituting in this value, and keeping the values for 𝐺, 𝑀, and π‘Ž the same as for the first part of the question, we find 𝐹=πΊπ‘€π‘šπ‘Ÿ=6.67Γ—10/β‹…Γ—10Γ—100(100000)=66.7.mkgskgkgmN

To two decimal places, we have 𝐹=66.70.N

The important thing to note here is that because the mass of object B is 20 times that of object A, the gravitational force object B experiences is also 20 times that felt by object A.

Part 4

In the final part of the question, we need to find the acceleration of object B toward the very large object. As we did for object A, we will use the equation π‘Ž=πΊπ‘€π‘Ÿ, with 𝐺=6.67Γ—10/β‹…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs, 𝑀=10kg, and π‘Ÿ=100000m. Since the mass of object B does not feature in this equation, these values are all the same as for the acceleration of object A that we calculated above, so we again find that π‘Ž=0.67/,ms to two decimal places.

Therefore, we have found that two objects of different masses that are at the same distance from some much larger mass experience the same acceleration due to gravity.

The fact that acceleration due to gravity does not depend on the mass of the object experiencing the acceleration means that when we consider any objects close to a large mass such as a moon or a planet, they all experience the same acceleration due to gravity.

Recall that the acceleration due to gravity on Earth’s surface is approximately 9.8 m/s2. We can now see where this value comes from, by using equation (4) with the mass of Earth, 𝑀=5.97Γ—10οŠͺkg. Any object or person on Earth’s surface is located a distance from Earth’s center of mass approximately equal to Earth’s radius, π‘Ÿ=6370km. We need to remember to convert the distance to metres, so that π‘Ÿ=6370000m. We can then use these values in the equation π‘Ž=πΊπ‘€π‘Ÿ, and we find π‘Ž=πΊπ‘€π‘Ÿ=6.67Γ—10/β‹…Γ—5.97Γ—10(6370000)=9.81/,οŠͺmkgskgms to two decimal places.

Strictly, the distance we use for π‘Ÿ should be the distance from Earth’s center to the center of mass of the object. However, Earth is so large that the height of the object’s center of mass above the surface for any object on or near the surface is insignificant compared to the radius of Earth.

We can demonstrate this by considering an object on the summit of Mount Everest, the highest mountain on Earth, which has a height of 8β€Žβ€‰β€Ž848 m. To calculate the acceleration due to gravity of this object, we would need to add the height of Everest to the radius of the Earth, so π‘Ÿ=6370000+8848=6378848m. If we use this value for π‘Ÿ, we find π‘Ž=πΊπ‘€π‘Ÿ=6.67Γ—10/β‹…Γ—5.97Γ—10(6378848)=9.79/,οŠͺmkgskgmms to two decimal places.

This is only slightly less than the acceleration experienced on Earth’s surface, and rounded to one decimal place the two values are the same, π‘Ž=9.8/ms. If this is all the accuracy we require, we can therefore use this value for the acceleration due to gravity anywhere on Earth, whether the object we are considering is at sea level or on top of the highest mountain.

Since the acceleration due to gravity on Earth is used so frequently, we give it the special symbol 𝑔. In general, the acceleration due to gravity on or near the surface of any large, spherical object is called the surface gravity.

Definition: Surface Gravity

The surface gravity of any large, spherical body such as a moon or planet is the acceleration due to gravity on or near the surface of that body.

We can calculate the surface gravity, π‘Ž, of any body using the equation π‘Ž=πΊπ‘€π‘Ÿ, where 𝐺=6.67Γ—10/β‹…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs is the universal gravitational constant, 𝑀 is the mass of the large body in kilograms, and π‘Ÿ is the radius in metres. Surface gravity is an acceleration and so has the unit of metres per second squared (m/s2).

Surface gravity is the same for all objects on or near the surface of the large body, regardless of the object’s mass.

In the case of Earth, surface gravity 𝑔=9.8/ms.

We can calculate the surface gravity on large, spherical bodies other than Earth if we have the mass and radius of the large body. We will look at the example of Ganymede, the largest moon of the planet Jupiter, below.

Example 2: Finding the Surface Gravity of a Moon

Ganymede is the largest moon in the solar system, with a mass of 1.48Γ—10 kg and a radius of 2β€Žβ€‰β€Ž630 km. What is the surface gravity on Ganymede? Give your answer to two decimal places.

Answer

As a moon, Ganymede is a large, roughly spherical body, so we can find the surface gravity using π‘Ž=πΊπ‘€π‘Ÿ, where the universal gravitational constant 𝐺=6.67Γ—10/β‹…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs, the mass of Ganymede 𝑀=1.48Γ—10kg, and the radius of Ganymede π‘Ÿ=2630km.

We first need to convert the radius to metres, so π‘Ÿ=2630=2630000kmm. We can then use these values in the equation for π‘Ž above, and we find π‘Ž=πΊπ‘€π‘Ÿ=6.67Γ—10/β‹…Γ—1.48Γ—10(2630000)=1.427…/.mkgskgms

Rounding to two decimal places, we have π‘Ž=1.43/.ms

Since the acceleration due to gravity that an object experiences does not depend on the mass of the object, we can say that it is a property of a given position in space close to a large spherical body, dependent only on the mass of the large body and the distance from its center of mass.

We can see from equation (4) that as distance from a large body’s center of mass increases, the acceleration due to gravity decreases. Often, objects are located sufficiently far from the surface of a large body that they experience significantly less acceleration due to gravity than they would at the surface. The general term for acceleration due to gravity at a specific point in space is the local acceleration due to gravity.

Surface gravity can also be referred to as the local acceleration due to gravity on the surface of a large, spherical body.

Next, we will look at an example of the local acceleration due to gravity of a satellite.

Example 3: Finding the Local Acceleration due to Gravity at a Position above Earth’s Surface

The International Space Station orbits Earth at a distance of 409 km above the surface. Earth has a mass of 5.97Γ—10οŠͺ kg and a radius of 6β€Žβ€‰β€Ž370 km. What is the local acceleration due to gravity at the height at which the International Space Station orbits? Give your answer to two decimal places.

Answer

The International Space Station is a crewed satellite that orbits Earth. It is sufficiently high above Earth’s surface that the local acceleration due to gravity is less than Earth’s surface gravity.

In order to calculate the local acceleration due to gravity at the height of the International Space Station’s orbit, we need to recall the equation for acceleration due to gravity, π‘Ž, which is π‘Ž=πΊπ‘€π‘Ÿ, where 𝐺=6.67Γ—10/β‹…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs is the universal gravitational constant, and Earth’s mass 𝑀=5.97Γ—10οŠͺkg is given in the question. For the distance from Earth’s center of mass, π‘Ÿ, we need to find the distance of the International Space Station from Earth’s center of mass.

We are given Earth’s radius, which we will call 𝑅=6370km, and the height of the International Space Station’s orbit above Earth’s surface, which we will call β„Ž=409km. A (not to scale) diagram is helpful here to visualize the setup.

From the example, we can see that the distance between the International Space Station and Earth’s center of mass π‘Ÿ=𝑅+β„Ž=6370+409=6779kmkm.

For the result to be an acceleration in metres per second squared (m/s2), we need π‘Ÿ in metres, so π‘Ÿ=6779000m.

We can then substitute this value, along with 𝐺 and 𝑀, into the equation for π‘Ž, to find π‘Ž=πΊπ‘€π‘Ÿ=6.67Γ—10/β‹…Γ—5.97Γ—10(6779000)=8.665…/.οŠͺmkgsms

Rounding to two decimal places, we have π‘Ž=8.67/.ms

As expected, this value is lower than Earth’s surface gravity 𝑔=9.81/ms, so the International Space Station experiences less acceleration due to gravity than an object on Earth’s surface.

Sometimes it is useful to compare the local acceleration due to gravity in two different positions. In the next example, we will look at the local acceleration due to gravity of a satellite relative to the surface gravity of Earth.

Example 4: Finding the Ratio of Local Acceleration due to Gravity at Two Positions

Geostationary satellites orbit Earth at a height of 35β€Žβ€‰β€Ž786 km above the equator. Earth has a mass of 5.97Γ—10οŠͺ kg and a radius of 6β€Žβ€‰β€Ž370 km. What is the ratio of the local acceleration due to gravity at the height of a geostationary satellite to that on the surface of Earth? Give your answer to three decimal places.

Answer

A geostationary satellite is one that remains above the same point on Earth’s surface throughout its orbit. We discuss their orbits in more detail in other explainers, but for now we are told that they orbit over the equator at a height of 53β€Žβ€‰β€Ž786 km.

In this question, we are finding the local acceleration due to gravity at that height relative to Earth’s surface gravity. If we call Earth’s radius 𝑅 and the height above Earth’s surface of the satellite’s orbit β„Ž, then the distance between the satellite and Earth’s center of mass π‘Ÿ=𝑅+β„Ž.

For the local acceleration due to gravity experienced by the satellite, we recall the equation for acceleration due to gravity: π‘Ž=πΊπ‘€π‘Ÿ, where 𝑀 is Earth’s mass, given in the question as 5.97Γ—10οŠͺ kg, and π‘Ÿ=𝑅+β„Ž is the distance between the satellite and Earth’s center of mass.

We can find Earth’s surface gravity from the same equation, using 𝑔 in place of π‘Ž and Earth’s radius, 𝑅, for the distance, so that 𝑔=𝐺𝑀𝑅.

We can then find the ratio π‘Žπ‘” by dividing the right-hand sides of these equations by each other. To make this easier, we will first find 1𝑔, which we can work out by turning the right-hand side of the equation upside down, so that 1𝑔=𝑅𝐺𝑀.

We can then use the fact that π‘Žπ‘”=π‘ŽΓ—1𝑔 and multiply the right-hand side of the equation for π‘Ž by the right-hand side of the equation for 1𝑔: π‘Žπ‘”=π‘ŽΓ—1𝑔=πΊπ‘€π‘Ÿπ‘…πΊπ‘€.

We can see that 𝐺𝑀 appears on both the top and bottom of the fraction on the right-hand side, meaning that they cancel out to leave π‘Žπ‘”=π‘…π‘Ÿ=ο€Όπ‘…π‘Ÿοˆ, where 𝑅 is Earth’s radius and π‘Ÿ is the height of the satellite above Earth’s center of mass.

We are given 𝑅=6370km and can calculate π‘Ÿ=𝑅+β„Ž=6370+53786=42156kmkmkm.

We would usually convert distances to metres before substituting values into an equation, but in this case we have a fraction with kilometres both on top and at the bottom, meaning that the units cancel each other out.

In this case, we can therefore substitute the values 𝑅=6370km and π‘Ÿ=42156km into the equation for π‘Žπ‘” to find π‘Žπ‘”=ο€Όπ‘…π‘Ÿοˆ=ο€½637042156=0.0228….kmkm

When rounded to three decimal places, we have π‘Žπ‘”=0.023.

As this quantity is a ratio of two quantities that both have the same units of acceleration, the resultant value has no units.

In the above example, we found the result that the ratio of the local acceleration due to gravity at two different distances from the same large spherical body, say acceleration π‘ŽοŽ at distance π‘Ÿ and π‘ŽοŒ± at distance 𝑅, depends only on the ratio of the distances, such that π‘Žπ‘Ž=ο€Όπ‘…π‘Ÿοˆ.

Remember that acceleration due to gravity is inversely proportional to the square of the distance, so if π‘ŽοŽ is on the top of the left-hand side, then π‘Ÿ must appear on the bottom of the right-hand side, and vice versa for π‘ŽοŒ± and 𝑅.

So far, when considering a small object in the vicinity of a large spherical body such as a moon or planet, we have only considered the acceleration due to gravity of the smaller object. In fact, both the small object and the large body experience acceleration due to the mass of the other.

If we have a large body of mass 𝑀, the acceleration, π‘Ž, of the body is related to the net force acting on it as 𝐹=π‘€π‘Ž, where the net force, 𝐹, is the gravitational force due to the presence of a small object of mass π‘š. The force is given by Newton’s law of gravitation, as in equation (2): 𝐹=πΊπ‘€π‘šπ‘Ÿ.

If we equate the right-hand sides of these equation, we find π‘€π‘Ž=πΊπ‘€π‘šπ‘Ÿ.

This time, we can divide both sides of the equation by 𝑀, which leaves π‘Ž=πΊπ‘šπ‘Ÿ.

So the large body experiences acceleration due to gravity from the mass of the small object in the same way that the small object is accelerated toward the large body.

In both cases, the acceleration acts along the line connecting the two objects’ centers of mass, acting to pull them closer together. The crucial thing to notice is that although the gravitational force is the same for both objects, the accelerations the two objects experience are not the same. The acceleration of an object is proportional to the mass of the other object. This means the object with lower mass experiences greater acceleration.

This can help us to understand why Earth does not appear to move because of the gravity of the people, buses, and satellites on its surface or in its vicinity. To see this in practice, we can look at an example of how Earth reacts to the gravity of a person.

Example 5: Finding Earth’s Acceleration due to the Gravity of a Person

A 75 kg person jumps off a 1 m high table onto the ground. If Earth has a radius of 6β€Žβ€‰β€Ž370 km, what is the acceleration of Earth due to the gravitational force between Earth and the person while they are in the air? Give your answer in scientific notation to two decimal places.

Answer

Here, we are considering the acceleration of Earth due to the gravity of a person 1 m above Earth’s surface. We can use a diagram to visualize the distances involved.

This diagram is, of course, not to scale, so in reality we do not need to consider the curvature of Earth; the surface is essentially flat on the scale of the table and the person.

To find Earth’s acceleration due to gravity, we need to recall the equation π‘Ž=πΊπ‘šπ‘Ÿ, where 𝐺=6.67Γ—10/β‹…οŠ±οŠ§οŠ§οŠ©οŠ¨mkgs is the universal gravitational constant, π‘š is the mass of the person, given in the question as π‘š=75kg, and π‘Ÿ is the distance between the person and the center of mass of Earth. In the diagram, we can see that if Earth’s radius is 𝑅=6370km, or 6β€Žβ€‰β€Ž370β€Žβ€‰β€Ž000 m, and the height of the person above the surface is 1 m, then π‘Ÿ=𝑅+β„Ž=6370001m.

In practice, β„Ž is so small compared to 𝑅 that we could just use π‘Ÿ=𝑅 and reach the same answer to the accuracy required.

If we substitute these values in, we find π‘Ž=πΊπ‘šπ‘Ÿ=6.67Γ—10/β‹…Γ—75(6370000)=1.232…×10/.mkgskgms

In scientific notation to two decimal places, this becomes π‘Ž=1.23Γ—10/.ms

We can see that the acceleration of Earth due to the gravity of a person is extremely small; this shows why Earth appears to be unaffected by the movement of people and buses on its surface.

In the same way that Earth is very slightly accelerated toward small objects in or near its surface, the Sun also experiences a small amount of acceleration toward Earth due to its gravity. The Sun has approximately a million times more mass than Earth, so this is much smaller than the acceleration Earth experiences toward the Sun, but it is enough to be measured as a very slight wobble in the Sun’s position. The acceleration of a star due to the gravity of its planets is one of the ways astronomers can find planets orbiting stars other than the Sun. Planets themselves are too small and faint to be seen, but the slight wobble of a star is enough to infer that it is there.

Key Points

  • The acceleration due to gravity, π‘Ž, of an object in the vicinity of another object of mass 𝑀 depends on 𝑀 and the distance between the two objects’ centers of mass such that π‘Ž=πΊπ‘€π‘Ÿ.
  • The acceleration due to gravity of an object does not depend on the mass of that object.
  • Surface gravity is the acceleration due to gravity on or near the surface of a large spherical body.
  • Earth’s surface gravity, π‘”β‰ˆ9.8/ms, is the acceleration due to gravity experienced by any object on or near Earth’s surface.
  • The general term for acceleration at a given distance from a large body is the local acceleration due to gravity.
  • Large bodies, such as Earth, also experience acceleration due to the gravity of small objects on its surface and nearby, but this is usually too small to be measured.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.