THE PRACTICE STUDIO
Go slowly. Ask why. Make it yours.
PRACTICE / Variation of gravity
Gravity at one planet radius of altitude
A spherical planet has radius R and surface gravity 12 m/s².
Find gravity at altitude R above its surface.
Organize the givens & assumptions
- gsurface = 12 m/s²
- h = R
Spherical planet, exterior inverse-square gravitational field; ignore rotation and other masses.
Study with a tutor prompt
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Problem: Gravity at one planet radius of altitude
A spherical planet has radius R and surface gravity 12 m/s².
Question: Find gravity at altitude R above its surface.
Givens:
- gsurface = 12 m/s²
- h = R
Assumptions: Spherical planet, exterior inverse-square gravitational field; ignore rotation and other masses.
Choices (LaTeX):
A. 3\ \mathrm{m\,s^{-2}}
B. 6\ \mathrm{m\,s^{-2}}
C. 12\ \mathrm{m\,s^{-2}}
D. 0\ \mathrm{m\,s^{-2}}
Ask for my approach first. Help me distinguish a physical misconception from an algebra error without revealing which choice is correct.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.Start with the smallest hint.
Work through it with me.
One decision at a time. Pause, predict, then unpack the algebra.
Distinguish altitude from distance to the centre in a gravitational-field model.
Choose a physical relation
Outside a spherical body, gravity depends on distance from its centre, not altitude alone.
Isolate what is asked
The centre-to-point distance is 2R, so the inverse-square field falls to one quarter.
Why is this step valid?
At altitude R the centre distance is 2R, so square the ratio R/(2R).
Substitute and interpret
Gravity is 3 m/s², one quarter of the surface value.
Doubling distance spreads the same gravitational flux over four times the spherical area.
TRY THE IDEA AGAIN
Would doubling the test object's mass change its gravitational acceleration here?
YOUR LEARNING NOTE
How does this idea feel now?
Confidence is a reflection, not proof of mastery.
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