🪐 Orbital Motion & Gravitation Simulator
GravitationWhat's happening? For a stable orbit, gravity must equal centripetal force. Too slow = falls inward. Too fast = escapes. This is how satellites stay in orbit.
Gravitation is a high-weightage topic — orbital motion, escape velocity, and Kepler's laws are asked every year.
JEE tests both the mathematical derivations (derive T² ∝ r³) and numerical applications (satellite orbits, escape velocity, gravitational potential energy).
Key formulas
Gravitation & Orbital Motion Formulas
Newton's gravity
F = GMm/r²
M = source mass, m = test mass, r = distance between centres.
Gravitational potential
V = -GM/r
Scalar. PE = -GMm/r. At infinity, V = 0. At surface, V = -GM/R.
Orbital velocity
v = √(GM/r)
Centripetal force = gravitational force. Stable orbit condition.
Kepler's 3rd law
T² ∝ r³, or T² = (4π²/GM)·r³
Time period squared proportional to radius cubed.
Escape velocity
vesc = √(2GM/r) = √2·vorb
Minimum speed to escape gravity — √2 times orbital velocity.
Energy of satellite
E = -GMm/2r
TE = KE + PE. KE = +GMm/2r, PE = -GMm/r, TE = -GMm/2r.
Step-by-step problem solving
How to solve gravitation problems in JEE
- Choose the right formula — use F = GMm/r² for forces, V = -GM/r for potential energies, v = √(GM/r) for orbits.
- Set up the balance — centripetal force = gravitational force for orbits: mv²/r = GMm/r².
- Handle height problems — if object is at height h above surface, r = R + h, not R.
- Use energy conservation — for escape velocity and projectile problems from planet surface.
- Remember: total energy is negative for bound systems — satellites have E < 0, escape requires E = 0.
Common JEE mistakes
What students get wrong
❌ Using r = R instead of R+h
If satellite is at height h, distance from centre = R + h. For low orbits (h << R), r ≈ R is OK, but be careful.
❌ Confusing escape and orbital velocity
vesc = √2 × vorb. If you calculate √(GM/r), you got orbital, not escape. The √2 factor is easy to miss.
❌ Positive potential energy
Gravitational PE = -GMm/r is ALWAYS negative (attractive force). Taking infinity as zero, any finite r gives negative PE.
❌ Assuming geostationary implies equatorial
A geostationary satellite must orbit in the equatorial plane. JEE sometimes asks for the latitude where a non-equatorial satellite can be geostationary — it must be the equator (0° latitude).
FAQs
Gravitation Questions
Why is gravitational PE negative?
We define PE = 0 at infinity. Since gravity is attractive, bringing a mass from infinity to distance r requires positive work from an external agent. But the gravitational potential energy stored is negative: PE = -GMm/r.
Can an object in circular orbit fall down?
No — if it's in a stable circular orbit, the centripetal force from gravity exactly balances the centrifugal effect. The object is in free fall continuously but never collides because the ground curves away at the same rate.
What happens if v > escape velocity?
The object escapes to infinity and still has positive kinetic energy there. Its excess speed (residual velocity) = √(v² - vesc²). The extra energy goes into KE at infinity.
Why does v = √(GM/r) come from force balance?
For a circular orbit, centripetal force mv²/r must equal gravitational force GMm/r². Setting them equal: mv²/r = GMm/r² → v² = GM/r → v = √(GM/r).