Why objects orbit

Track position, velocity, and acceleration separately

Start the demonstration at Reset. The two objects share a starting position on the right side of the circle and the same upward velocity. One follows the curved orbit; the other follows the dashed straight path with gravity removed. This comparison holds the initial motion fixed and changes the force. It does not compare a moving spacecraft with a stationary object, and it does not show a rocket launching from Earth's surface.

Use Next step to advance once. The straight-path object keeps moving upward. The orbiting object has also moved toward the left, because gravity has changed its velocity toward Earth. Its green velocity arrow remains tangent to the circle. The amber gravity arrow points toward the center, not along the path already traveled. An acceleration arrow tells you how velocity is changing now, not where the spacecraft will be after a long interval.

Advance through all four steps. At the endpoint, the spacecraft is above Earth and its velocity points left. The gravity-free object is still on the vertical dashed line. The orbiting object's speed has not decreased to zero at the top. The top of this drawing is not the highest point of a vertically thrown ball. It is another point at the same distance from Earth's center, with sideways motion relative to the local radius.

To compare velocities at the start and end, draw their arrows from a common origin. The first points up and the second left. Their equal lengths represent equal speeds, but their tips differ, so their difference is not zero. That difference divided by elapsed time gives average acceleration over the interval. It is not the same as the instantaneous gravity arrow at either endpoint. Gravity changes direction continuously as the spacecraft moves around Earth.

The diagram uses four equal teaching intervals to show a quarter of a circular orbit. Its percentages describe progress through that demonstration, not the fraction of a full orbit completed. At 100 percent the spacecraft has turned through ninety degrees, not three hundred sixty. Playback seconds are not orbital seconds. Earth is enlarged, so measuring the drawn gap between its surface and the orbit cannot recover an altitude.

Why does an inward force not steadily increase speed here? At each instant in a circle, velocity is perpendicular to gravity. Over a very short interval the velocity turns toward the center. The radial distance stays fixed, and the force supplies the turning required by that motion. On an ellipse, velocity can also have an inward or outward part. Gravity then changes speed as well as direction. Constant speed is a property of this circular example, not of every orbit.

Do not add a second outward force merely because the radius stays constant. In the nonrotating frame used here, gravity is the net force and the craft accelerates inward. Constant distance from the center does not mean zero acceleration. A rotating coordinate system can introduce an apparent outward term to describe the same motion, but that is a different bookkeeping system. Mixing its apparent force into the nonrotating calculation would count incompatible descriptions together.

The same distinction explains a floating astronaut. A bathroom scale measures the supporting contact force exerted through its surface. On the ground, that force prevents you from accelerating downward. In an ideal freely falling cabin, you and the scale accelerate together, and that support force can be nearly zero even while gravity acts strongly. Real spacecraft have small disturbances and gravity differences across their size, so actual weightlessness is approximate.

Derive the speed and check what the numbers mean

The expression for inward acceleration can be understood without assuming that circular motion has a hidden outward force. Over a short interval, the radius vector turns through a small angle. The velocity vector turns through the same angle because it stays perpendicular to the radius. The short distance traveled is approximately speed times elapsed time. Dividing that distance by radius gives the small angle in radians.

For the corresponding small turn in velocity, the magnitude of the velocity change is approximately speed times that angle. Substitute the previous expression for the angle and divide by elapsed time. The result approaches speed squared divided by radius as the interval becomes smaller. This is the acceleration needed for the chosen circle and speed. It describes the motion; gravity must supply it if there is no other force.

Now equate the required acceleration with the available gravitational acceleration. Both sides contain a factor involving radius. Multiplying both sides by radius leaves speed squared equal to mu divided by radius. Taking the positive square root gives the speed magnitude. A positive magnitude does not choose a direction of travel. Clockwise and counterclockwise tangential velocities can both produce circular motion in this ideal spherical field.

Work through the 400 km example in stages. First add the mean Earth radius, 6,371 km, to the altitude, giving 6,771 km from the center. Divide 398,600.4418 by 6,771 to obtain approximately 58.87 square kilometers per square second. Its square root is about 7.673 kilometers per second. Multiplying by 3,600 seconds per hour gives about 27,620 kilometers per hour. The reported 27,600 is a rounded value, not a different physical prediction.

The unit check follows the algebra. Mu has units of cubic kilometers per square second. Dividing by kilometers leaves square kilometers per square second. Taking a square root leaves kilometers per second. If you insert radius in meters while keeping mu in kilometer units, the arithmetic may still produce a number, but the units no longer cancel as assumed. Convert every length consistently before interpreting the result.

Check the acceleration independently at the same radius. Mu divided by radius squared gives about 0.00869 kilometers per square second, or 8.69 meters per square second. Alternatively, square the calculated circular speed and divide by radius. The two methods should agree because that equality defined the circular solution. This is a useful calculation check, not an independent experimental measurement of Earth's gravity.

A radius comparison gives a prediction before any calculator work. Doubling the radius makes gravitational acceleration one quarter as large. Circular speed falls only by the square root of two, to about 0.707 of its former value. Substituting that speed into speed squared divided by radius gives half the numerator and twice the denominator, again one quarter of the original acceleration. The larger orbit requires less inward acceleration because both speed and curvature change.

Doubling altitude is a different comparison. Moving from 400 km to 800 km altitude changes radius from 6,771 km to 7,171 km, not to 13,542 km. The circular speed changes from about 7.673 to 7.456 km/s, a small decrease rather than the factor 0.707. Always state whether a distance is measured from the surface or from the center before applying a ratio argument.

Decide whether a circular calculation answers the question

A prediction needs a starting state as well as a force law. Here a state means position and velocity at a specified time. Knowing altitude alone gives a sphere of possible positions, not one spacecraft location. Knowing speed alone leaves many possible velocity directions. The circular calculation adds a restrictive condition: the velocity must be perpendicular to the radius and have the magnitude needed to keep that radius constant.

Consider two craft at the same 400 km altitude with the same speed of about 7.673 km/s. Give one a tangential velocity and the other a velocity directed straight toward Earth's center. The first fits the circular model. The second does not acquire sideways motion merely because its speed happens to equal the circular value. In the ideal central field it heads inward, encountering the atmosphere and planet rather than completing the circle.

Likewise, switching an engine off does not erase an existing sideways velocity. In the gravity-free comparison, motion continues in a straight line. In the gravity-on comparison, motion continues along a curved path. A coasting spacecraft needs no continuous forward force to preserve its motion in the absence of resistance. A car on a road usually needs power because resistance removes mechanical energy; that familiar experience cannot simply be transferred to an ideal vacuum orbit.

The negligible-spacecraft-mass assumption also matters. Both bodies attract each other. In a precise two-body treatment they orbit their common center of mass, and their relative acceleration involves the sum of their masses. Earth-centered motion with a fixed Earth and mu equal to Earth's gravitational parameter is an excellent approximation when spacecraft mass is tiny compared with Earth's. It would need revision for two bodies of comparable mass.

Changing the coordinate system changes how motion is described. In the nonrotating Earth-centered picture, a circular craft follows a circle at the calculated speed. A point on Earth's surface is rotating underneath that path. Speed relative to the ground therefore need not equal the Earth-centered speed. Comparing observations requires a common reference frame; subtracting numbers reported in different frames without accounting for their directions can give a meaningless result.

Before trusting a calculated circle, check whether it lies outside the planet and whether neglected effects matter over the interval of interest. Thin upper atmosphere can gradually alter a low orbit even when the instantaneous gravitational calculation is accurate. An uneven gravitational field can alter the orientation of a path over time. These effects do not invalidate Newton's laws. They show that the list of forces in the simple model is incomplete for that prediction.

The useful result of this lesson is therefore conditional. Given a spherical Earth, a small coasting craft, negligible drag, and the appropriate tangential starting velocity, the calculation predicts circular motion and its speed. A real launch must establish that state while passing through an atmosphere and consuming propellant. Reaching the stated altitude is only one requirement. Later lessons relax the circle assumption to explain how changing velocity connects different orbital paths.