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Gravitational Force Calculator

The Gravitational Force Calculator computes the gravitational force between two masses using Newton's law of universal gravitation. Enter any three values to solve for the fourth.

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    What is Gravitational Force?

    Gravitational force is a fundamental physical interaction that causes mutual attraction between all objects with mass or energy. It is the weakest of the four fundamental forces of nature, yet it dominates at macroscopic and cosmic scales [1].

    This force governs the motion of planets, stars, and galaxies across the universe. On Earth, it gives weight to physical objects and drives the ocean tides through continuous lunar interaction.

    Newton's Law of Universal Gravitation

    Isaac Newton formulated the law of universal gravitation in 1687. The law states that every point mass attracts every other point mass in the universe with a force proportional to the product of their masses and inversely proportional to the square of the distance between them [2].

    The primary equation is expressed mathematically as follows. This relationship allows physicists to predict orbital mechanics and celestial behavior with high precision.

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

    The variables in this equation are defined by their standard International System of Units (SI). The table below outlines the exact parameters used in the calculation.

    Symbol Quantity SI Unit Description
    $F$ Gravitational Force Newton (N) The attractive pull between the two masses.
    $G$ Gravitational Constant $\text{N}\cdot\text{m}^2/\text{kg}^2$ The universal proportionality constant.
    $m_1$ Mass 1 Kilogram (kg) The mass of the first object.
    $m_2$ Mass 2 Kilogram (kg) The mass of the second object.
    $r$ Distance Meter (m) The distance between the centers of the masses.

    How to Calculate Gravitational Force (Step-by-Step)

    Calculating gravitational force requires identifying the masses of both objects and the distance between their centers. The universal gravitational constant is then applied to find the resulting force.

    Example 1: The Force Between Two Astronauts

    Two astronauts, each with a mass of 70 kilograms, are floating in space exactly 1 meter apart during a spacewalk. Calculate the gravitational force they exert on each other.

    1. Identify the known values: $m_1 = 70 \text{ kg}$, $m_2 = 70 \text{ kg}$, $r = 1 \text{ m}$, and $G = 6.674 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$.
    2. Substitute the values into the primary equation: $$ F = (6.674 \times 10^{-11}) \frac{70 \times 70}{1^2} $$
    3. Multiply the masses in the numerator to find the intermediate value: $$ F = (6.674 \times 10^{-11}) \frac{4900}{1} $$
    4. Calculate the final microscopic force: $$ F = 3.27 \times 10^{-7} \text{ N} $$

    This microscopic value explains why humans do not physically feel the gravitational pull of other people. The force is entirely overwhelmed by Earth's gravity and physical friction.

    Example 2: The Earth and the Moon

    An astrophysicist needs to verify the gravitational binding force of the Earth-Moon system. Given the Earth's mass is $5.972 \times 10^{24}$ kg, the Moon's mass is $7.342 \times 10^{22}$ kg, and their average center-to-center distance is $3.844 \times 10^8$ meters, determine the gravitational force between them.

    1. Substitute the massive astronomical values into the primary equation: $$ F = (6.674 \times 10^{-11}) \frac{(5.972 \times 10^{24}) \times (7.342 \times 10^{22})}{(3.844 \times 10^8)^2} $$
    2. Multiply the masses and square the distance to establish the intermediate fractions: $$ F = (6.674 \times 10^{-11}) \frac{4.385 \times 10^{47}}{1.478 \times 10^{17}} $$
    3. Divide the numerator by the denominator and apply the constant: $$ F = (6.674 \times 10^{-11}) \times (2.967 \times 10^{30}) $$
    4. Calculate the final macroscopic force: $$ F \approx 1.98 \times 10^{20} \text{ N} $$

    This massive attractive force is exactly what keeps the Moon in a stable orbit around the Earth. It demonstrates why scientific notation is strictly necessary for astronomical calculations.

    Gravity Equations and Derived Formulas

    The primary formula can be algebraically rearranged to solve for any unknown variable. This flexibility allows scientists to determine planetary masses or orbital distances using observed gravitational effects.

    How to Calculate Mass from Gravitational Force?

    When the gravitational force, distance, and one mass are known, the unknown mass can be isolated. The derived formula is expressed as follows.

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

    Example 1: Estimating the Mass of the Earth

    A 1 kg object at the surface of the Earth experiences a gravitational force of 9.8 Newtons. Given the Earth's radius is $6.371 \times 10^6$ meters, calculate the mass of the Earth.

    1. Substitute the known surface gravity and planetary radius into the derived mass formula: $$ m_1 = \frac{9.8 \times (6.371 \times 10^6)^2}{(6.674 \times 10^{-11}) \times 1} $$
    2. Square the radius and multiply by the force to find the numerator: $$ m_1 = \frac{9.8 \times (4.059 \times 10^{13})}{6.674 \times 10^{-11}} $$
    3. Calculate the final numerator value: $$ m_1 = \frac{3.978 \times 10^{14}}{6.674 \times 10^{-11}} $$
    4. Divide to isolate the Earth's mass: $$ m_1 \approx 5.97 \times 10^{24} \text{ kg} $$

    Example 2: Finding an Exoplanet's Mass

    A space probe orbits an unknown exoplanet at a distance of $2 \times 10^7$ meters. The probe, with a mass of 100 kg, experiences a gravitational pull of 500 Newtons. Calculate the mass of the exoplanet.

    1. Insert the satellite parameters and the measured force into the mass equation: $$ m_1 = \frac{500 \times (2 \times 10^7)^2}{(6.674 \times 10^{-11}) \times 100} $$
    2. Square the distance and multiply by the force: $$ m_1 = \frac{500 \times (4 \times 10^{14})}{6.674 \times 10^{-9}} $$
    3. Calculate the final numerator value: $$ m_1 = \frac{2 \times 10^{17}}{6.674 \times 10^{-9}} $$
    4. Divide to reveal the planet's mass: $$ m_1 \approx 1.49 \times 10^{25} \text{ kg} $$

    Example 3: Calculating a Secondary Star's Mass

    In a binary star system, the primary star exerts a force of $1.5 \times 10^{18}$ Newtons on its companion. The distance between them is $5 \times 10^{11}$ meters, and the primary star's mass is $2 \times 10^{30}$ kilograms. Calculate the mass of the secondary star.

    1. Apply the mass derivation formula using the binary system parameters: $$ m_1 = \frac{(1.5 \times 10^{18}) \times (5 \times 10^{11})^2}{(6.674 \times 10^{-11}) \times (2 \times 10^{30})} $$
    2. Square the separation distance and multiply by the force: $$ m_1 = \frac{(1.5 \times 10^{18}) \times (2.5 \times 10^{23})}{1.335 \times 10^{20}} $$
    3. Calculate the final numerator value: $$ m_1 = \frac{3.75 \times 10^{41}}{1.335 \times 10^{20}} $$
    4. Divide to find the secondary mass: $$ m_1 \approx 1.25 \times 10^{30} \text{ kg} $$

    How to Calculate Distance from Gravitational Force?

    If the masses and the gravitational force are known, the separation distance can be determined. The formula is rearranged using a square root.

    $$ r = \sqrt{\frac{G \cdot m_1 \cdot m_2}{F}} $$

    Example 1: Determining Geostationary Orbit Radius

    A communications satellite requires a gravitational force of 236 Newtons to maintain a geostationary orbit around the Earth. Given the satellite's mass is 1,000 kg and the Earth's mass is $5.972 \times 10^{24}$ kg, calculate the required orbital radius.

    1. Substitute the Earth's mass and the required centripetal force into the distance formula: $$ r = \sqrt{\frac{(6.674 \times 10^{-11}) \times (5.972 \times 10^{24}) \times 1000}{236}} $$
    2. Multiply the constant by both masses to establish the numerator: $$ r = \sqrt{\frac{3.986 \times 10^{17}}{236}} $$
    3. Divide the numerator by the required force: $$ r = \sqrt{1.689 \times 10^{15}} $$
    4. Take the square root of the result to find the orbital radius: $$ r \approx 4.22 \times 10^7 \text{ m} $$

    Example 2: Cavendish Experiment Separation

    In a laboratory Cavendish experiment, two 10 kg lead spheres attract each other with a measured force of $6.67 \times 10^{-10}$ Newtons. Calculate the distance between their centers.

    1. Insert the known lead sphere masses and the measured laboratory force into the equation: $$ r = \sqrt{\frac{(6.674 \times 10^{-11}) \times 10 \times 10}{6.67 \times 10^{-10}}} $$
    2. Multiply the constant by the two masses: $$ r = \sqrt{\frac{6.674 \times 10^{-9}}{6.67 \times 10^{-10}}} $$
    3. Divide the numerator by the tiny measured force: $$ r = \sqrt{10.006} $$
    4. Calculate the square root to find the exact separation distance: $$ r \approx 1 \text{ m} $$

    Example 3: Asteroid Belt Distance

    Two asteroids in the main belt, each with a mass of $1 \times 10^{15}$ kilograms, exert a mutual gravitational force of 10 Newtons. Calculate the distance separating them.

    1. Apply the square root formula to the known asteroid masses and their mutual force: $$ r = \sqrt{\frac{(6.674 \times 10^{-11}) \times (1 \times 10^{15}) \times (1 \times 10^{15})}{10}} $$
    2. Multiply the gravitational constant by the product of the two massive asteroids: $$ r = \sqrt{\frac{6.674 \times 10^{19}}{10}} $$
    3. Divide by the force to isolate the squared distance: $$ r = \sqrt{6.674 \times 10^{18}} $$
    4. Extract the square root to find their separation: $$ r \approx 2.58 \times 10^7 \text{ m} $$

    How to Calculate Gravitational Field Strength (Acceleration)?

    Gravitational field strength, often denoted as little $g$, represents the force exerted per unit of mass. The formula is derived by dividing the primary force equation by the test mass.

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

    Example 1: Surface Gravity of Mars

    Calculate the acceleration due to gravity on the surface of Mars. The planet has a mass of $6.39 \times 10^{23}$ kg and a mean radius of $3.39 \times 10^6$ meters.

    1. Substitute the planetary mass and mean radius into the field strength formula: $$ g = \frac{(6.674 \times 10^{-11}) \times (6.39 \times 10^{23})}{(3.39 \times 10^6)^2} $$
    2. Multiply the gravitational constant by the mass of Mars: $$ g = \frac{4.265 \times 10^{13}}{(3.39 \times 10^6)^2} $$
    3. Square the radius to find the denominator: $$ g = \frac{4.265 \times 10^{13}}{1.149 \times 10^{13}} $$
    4. Divide to find the surface acceleration: $$ g \approx 3.71 \text{ m/s}^2 $$

    Example 2: Gravity at the ISS Altitude

    The International Space Station orbits at an altitude where its total distance from the Earth's center is $6.771 \times 10^6$ meters. Calculate the gravitational field strength at this specific altitude.

    1. Use the Earth's mass and the total orbital radius in the field strength equation: $$ g = \frac{(6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{(6.771 \times 10^6)^2} $$
    2. Multiply the constant by the Earth's mass to establish the numerator: $$ g = \frac{3.986 \times 10^{14}}{(6.771 \times 10^6)^2} $$
    3. Square the total distance to find the denominator: $$ g = \frac{3.986 \times 10^{14}}{4.585 \times 10^{13}} $$
    4. Divide to calculate the local gravity: $$ g \approx 8.66 \text{ m/s}^2 $$

    Example 3: Comparing Jupiter and Earth

    Compare the surface gravity of Jupiter to that of the Earth. Jupiter has a mass of $1.898 \times 10^{27}$ kg and a radius of $6.99 \times 10^7$ meters.

    1. Apply the field strength formula using the mass and radius of Jupiter: $$ g = \frac{(6.674 \times 10^{-11}) \times (1.898 \times 10^{27})}{(6.99 \times 10^7)^2} $$
    2. Multiply the constant by the mass to establish the numerator: $$ g = \frac{1.267 \times 10^{17}}{(6.99 \times 10^7)^2} $$
    3. Square the radius to find the denominator: $$ g = \frac{1.267 \times 10^{17}}{4.886 \times 10^{15}} $$
    4. Divide to calculate the surface gravity: $$ g \approx 24.79 \text{ m/s}^2 $$

    How to Use This Gravitational Force Calculator

    This tool simplifies complex astronomical and terrestrial calculations into three straightforward steps. Follow this sequence to get accurate results for any scenario.

    1. Select the specific variable you need to calculate from the dropdown menu at the top of the widget.
    2. Input your known values into the corresponding fields and select the appropriate units from the dynamic dropdowns.
    3. View your primary result in the designated output field, along with the equivalent scientific notation displayed directly beneath it.

    The internal engine automatically converts mixed units into base metric standards. This ensures complete accuracy whether you are combining Solar Masses with Astronomical Units.

    What is the Gravitational Constant (G)?

    The gravitational constant, denoted as $G$, is an empirical physical constant used to calculate the gravitational attraction between two bodies. Its currently accepted value is approximately $6.67430 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2$ [3].

    This constant was first accurately measured by Henry Cavendish in 1798. He used a highly sensitive torsion balance to measure the tiny gravitational attraction between lead spheres [4].

    It is critical to distinguish this universal constant from the local acceleration due to gravity. The universal constant $G$ applies everywhere in the cosmos, while local gravity $g$ varies depending on the mass and radius of the specific celestial body.

    Gravitational Force Units

    The standard International System of Units (SI) unit for force is the Newton. One Newton is defined as the force required to accelerate a one-kilogram mass at a rate of one meter per second squared [5].

    Different scientific and engineering disciplines sometimes utilize alternative units for convenience. The calculator automatically handles the complex conversions between all supported units.

    Unit Name Symbol Equivalent in Newtons Common Use Case
    Newton N 1 Standard SI unit for all physics.
    Kilonewton kN 1,000 Structural engineering and thrust.
    Meganewton MN 1,000,000 Rocket engine specifications.
    Dyne dyn 0.00001 CGS system and astrophysics.
    Pound-force lbf 4.44822 Imperial system applications.
    Kilogram-force kgf 9.80665 Legacy metric engineering.

    How Does Distance Affect Gravitational Force?

    Gravitational force obeys the inverse-square law. This means the force is inversely proportional to the square of the distance between the two objects [6].

    If you double the distance between two masses, the gravitational force drops to one-quarter of its original value. If you triple the distance, the force drops to one-ninth.

    This rapid drop-off explains why gravity is highly localized to massive bodies. Despite having an infinite theoretical range, the force becomes negligible at vast interstellar distances.

    Distance Multiplier Force Fraction Force Percentage
    1x (Baseline) 1 / 1 100%
    2x 1 / 4 25%
    3x 1 / 9 11.11%
    4x 1 / 16 6.25%
    5x 1 / 25 4.00%
    10x 1 / 100 1.00%

    Frequently Asked Questions

    Can I use different units for mass, force, and distance?

    Yes, the calculator allows you to mix and match units across all input fields. The internal engine automatically converts them to base metric standards before calculating the final result.

    What is the difference between gravitational force and weight?

    Weight is a specific type of gravitational force exerted by a large celestial body on a smaller object near its surface [7]. Gravitational force is the broader universal term for the attraction between any two masses in space.

    Does the medium between objects affect gravitational force?

    No, gravity acts through a complete vacuum and is entirely unaffected by the medium between the objects. Unlike electromagnetic forces, there is no known material that can block or shield gravitational fields.

    Why is the gravitational force between everyday objects unnoticeable?

    The gravitational constant is incredibly small, requiring at least one object to have a planetary-scale mass for the resulting force to be noticeable. The microscopic pull between humans or everyday items is completely overwhelmed by Earth's gravity and physical friction.

    Can gravity be blocked or shielded?

    No known material can block or shield gravitational fields from passing through them. Gravity interacts with all mass and energy uniformly, making it impossible to create a gravitational shadow.

    How do astronauts experience weightlessness if gravity is still strong in orbit?

    Astronauts experience weightlessness because their spacecraft and everything inside it are in a state of continuous freefall around the Earth [8]. They are falling at the exact same rate as their surroundings, creating the sensation of zero gravity.

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    For Further Reading

    The concepts and formulas detailed on this page are grounded in established classical mechanics. The following authoritative resources provide deeper mathematical derivations and historical context.

    1. Newton's Law of Universal Gravitation (Encyclopædia Britannica)
    2. Gravitation (OpenStax University Physics via LibreTexts)
    3. Newtonian Constant of Gravitation (NIST Reference on Constants)
    4. How Cavendish Weighed the Earth (Science History Institute)
    5. SI Base Units (Bureau International des Poids et Mesures)
    6. Inverse Square Law (HyperPhysics, Georgia State University)
    7. What is Weight? (NASA Science for Students)
    8. What is Microgravity? (NASA Science for Students)