Explain gravity and acceleration in orbit

Begin with the direction of motion

Draw a circle around Earth and put a spacecraft at the rightmost point. For counterclockwise motion, draw its velocity arrow upward, tangent to the circle. Draw its acceleration arrow leftward, toward Earth’s center. The two arrows answer different questions: velocity describes motion now; acceleration describes how velocity changes. An inward acceleration bends the path without requiring the speed to decrease.

Move your paper spacecraft to the top of the circle. Its velocity now points left and its acceleration points down. Repeat at the other two quarter-turn positions. The velocity arrows have equal lengths but different directions. Because velocity includes direction, constant speed around a circle is accelerated motion.

NASA describes an orbit as continued free fall under gravity. Free fall does not mean the gravitational field has disappeared. NASA gravity and mechanics. This lab isolates a circular case, with no engine burn or atmospheric force. Gravity supplies the inward acceleration rather than being cancelled by an outward force in the displayed inertial frame.

Measure radius from the center

The radius control is expressed in Earth radii. The model uses a spherical Earth of radius R = 6371 km and gravitational parameter μ = GM = 398600 km³/s². Here G is the gravitational constant and M is Earth’s mass; their product sets the field strength. These rounded parameters follow the NASA Earth fact sheet.

If the control is 2, the orbital radius is r = 2R = 12742 km, measured from Earth’s center. Altitude above the spherical surface is h = r − R = 6371 km. Radius and altitude are not interchangeable. At control 1.1, radius is 7008.1 km but altitude is only 637.1 km.

Make a paper ledger for controls 1.1, 2 and 4. Write both radius and altitude before looking at acceleration. This prevents a common inverse-square error: inserting altitude where the formula requires center-to-center distance.

Calculate how gravity changes with distance

The acceleration magnitude in this model is g = 1000μ/r² in m/s², with r entered in kilometers. The factor 1000 converts the resulting km/s² to m/s². An inverse square means that doubling the radius makes the denominator four times larger.

At r = 12742 km, g is about 2.4551 m/s². At r = 25484 km, it is about 0.6138 m/s², one quarter as large. Gravity is weaker there, but it is not zero. A freely falling object can remain strongly accelerated even when no floor supplies a support force.

Hold the elapsed time at zero while comparing these radii. Then hold radius at 2R and change elapsed time. The acceleration magnitude stays constant along that circular orbit while its direction follows the spacecraft toward the center. State which comparison changes the field magnitude and which changes only the displayed direction.

State the mass assumption

The orbiting object’s mass is treated as negligible compared with Earth’s mass. Within this approximation, the object’s own mass does not appear in the acceleration or circular-speed expressions. This is not a model of two comparable stars orbiting their shared center of mass.

The displayed path is selected as circular from the start. Changing radius chooses a different circular solution and its matching speed; it does not calculate a spacecraft firing an engine to transfer between those circles. The arrows illustrate the chosen solution rather than a record from a tracking instrument.

Transfer to an unfamiliar ratio

A second circular orbit has three times the first orbit’s center-based radius. If the first acceleration is 2.7 m/s², what is the second acceleration? Explain why tripling altitude would not generally produce the same ratio.

Check the orbital-gravity transfer

The acceleration is 2.7/3² = 0.3 m/s². Tripling altitude changes r from R + h to R + 3h, which is not generally three times R + h. The inverse-square relationship uses radius measured from Earth's center.