Questions: Gravitation

Q1. State Newton's law of universal gravitation and write it in vector form.

Answer: Every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them. In vector form the force on mass one due to mass two is minus G m1 m2 over r squared times the unit vector from one to two, where the minus sign shows that the force is attractive and points from one toward two. The constant G is universal and does not depend on the medium or on the masses.

Hinglish: Har particle doosre particle ko apni taraf attract karta hai, force masses ke product ke proportional aur distance ke square ke inverse proportional hoti hai. Vector form mein minus sign attractive nature dikhata hai. G universal constant hai, medium ya masses par depend nahi karta.

Q2. State Kepler's three laws.

Answer: The first law says each planet moves in an ellipse with the sun at one focus. The second law says that the area swept out by the line from the sun to the planet grows at a constant rate. The third law says that the square of the period varies as the cube of the semi-major axis. The second law is angular momentum conservation, and the third law follows from the inverse square force law.

Hinglish: Pehla law: planet ka orbit ellipse hota hai aur sun ek focus par. Doosra: sun se planet tak ki line barabar samay mein barabar area sweeps karti hai. Teesra: period ka square semi-major axis ke cube ke proportional. Doosra law angular momentum conservation hai, aur teesra inverse square force se nikalta hai.

Q3. Derive Kepler's third law for a circular orbit.

Answer: For a circular orbit the gravitational force supplies the centripetal force, so G M m over r squared equals m v squared over r. This gives v squared equal to G M over r. The period is the circumference divided by the speed, so T is 2 pi r over v, and squaring it gives T squared equal to 4 pi squared r cubed over G M. So the square of the period is proportional to the cube of the radius, with the constant depending only on the central mass.

Hinglish: Circular orbit mein gravitation force centripetal force deta hai: GMm/r^2 = mv^2/r, isliye v^2 = GM/r. Period T = 2 pi r / v, square karne par T^2 = 4 pi^2 r^3/(GM). Isliye period ka square radius ke cube ke proportional hai.

Q4. Why does the acceleration due to gravity decrease with height?

Answer: The gravitational force falls as the inverse square of the distance from the centre of the Earth. At a height h the distance becomes R plus h, so the acceleration is G M over the square of R plus h, which is smaller than the surface value. For heights much smaller than the radius the decrease is about 2h over R as a fraction. At very large heights, such as the Moon's orbit, the decrease is large and the value is a small fraction of the surface value.

Hinglish: Gravity force centre se distance ke square ke inverse proportional hai. Height h par distance R + h ho jaata hai, isliye acceleration GM/(R+h)^2 chhota ho jaata hai. Chhoti heights ke liye fractional decrease lagbhag 2h/R hota hai. Bahut badi height par g kaafi chhota ho jaata hai.

Q5. Define gravitational potential and write its expression for a point mass.

Answer: The gravitational potential at a point is the work done per unit mass by an external agent in bringing a small test mass from infinity to that point without acceleration. Taking the potential as zero at infinity, the value at distance r from a point mass M is minus G M over r. It is negative because gravity is attractive and the external agent must do positive work to move the mass away. Potential is a scalar, so potentials of several masses add directly.

Hinglish: Gravitational potential woh work per unit mass hai jo external agent infinity se point tak laane mein karta hai. Infinity par zero maan kar point mass M se distance r par V = -GM/r hoti hai. Negative isliye kyunki gravity attractive hai. Potential scalar hai, isliye seedha add hota hai.

Q6. What is escape speed? Derive its expression.

Answer: The escape speed is the minimum speed with which a body must be projected from the surface so that it reaches infinity and never returns. Set the total energy at the surface to zero, because at infinity the body just comes to rest with zero potential energy. The kinetic energy at the surface equals the magnitude of the potential energy, so half m v squared equals G M m over R. This gives v equal to the square root of twice G M over R, about 11.2 km per second for the Earth, independent of the mass and direction.

Hinglish: Escape speed minimum speed hai jisse body infinity tak jaakar wapas na aaye. Surface par total energy zero rakho: 1/2 m v^2 = GMm/R, isliye v = root(2GM/R), Earth ke liye lagbhag 11.2 km/s. Yeh mass aur direction par depend nahi karti.

Q7. Why is an astronaut in an orbiting spacecraft weightless?

Answer: In orbit the spacecraft and everything inside it fall toward the Earth with the same acceleration, so the astronaut and the floor fall together and the normal reaction between them is zero. The astronaut's weight is not zero, because gravity still acts; what is zero is the apparent weight, which is the normal reaction. The effect is the same as in a freely falling lift, and it is called weightlessness in free fall.

Hinglish: Orbit mein spacecraft aur astronaut dono same acceleration se girte hain, isliye floor aur astronaut ke beech normal reaction zero ho jaata hai. Gravity to lagti hai, weight zero nahi hota; apparent weight zero hota hai. Yeh free fall wale lift jaisa hi effect hai.

Q8. Compare the escape speed and the orbital speed of a low satellite.

Answer: The orbital speed of a satellite close to the Earth's surface is the square root of G M over R, and the escape speed is the square root of twice G M over R. The ratio is root two, so the escape speed is about 1.41 times the orbital speed. Numerically the orbital speed is about 7.9 km per second and the escape speed about 11.2 km per second. A body launched at the orbital speed circles the Earth, and one launched at the escape speed leaves it.

Hinglish: Orbital speed root(GM/R) aur escape speed root(2GM/R) hoti hai, isliye ratio root 2 hai. Orbital speed lagbhag 7.9 km/s aur escape speed lagbhag 11.2 km/s hoti hai. Orbital speed par body Earth ke around ghoomti hai, escape speed par nikal jaati hai.

Q9. Find the height at which the acceleration due to gravity is one fourth of its value at the surface.

Answer: The ratio of the value at height h to the value at the surface is R squared over the square of R plus h. Setting this equal to one fourth gives R over R plus h equal to one half, so R plus h is twice R and h equals R. The height is one Earth radius, about 6400 km. A check is that doubling the distance from the centre reduces the inverse square to one quarter, which matches.

Hinglish: Height h par g_h/g = R^2/(R+h)^2. Isko 1/4 ke barabar rakhne se R/(R+h) = 1/2, matlab R + h = 2R aur h = R, yani lagbhag 6400 km. Centre se distance double hone par inverse square one fourth ho jaata hai, jo check hai.

Q10. What is the total energy of a satellite in a circular orbit?

Answer: The kinetic energy is half the magnitude of the potential energy, so with potential energy minus G M m over r the kinetic energy is G M m over 2r. The total is the sum, which is minus G M m over 2r, a negative quantity. The negative sign means the satellite is bound to the Earth. The energy needed to escape from the orbit is the negative of the total energy, that is G M m over 2r.

Hinglish: Kinetic energy potential energy ke magnitude ka aadha hota hai: K = GMm/2r, aur U = -GMm/r. Total E = -GMm/2r, jo negative hai, matlab satellite bound hai. Orbit se escape karne ke liye GMm/2r energy chahiye, jo total energy ka negative hai.

Quiz below: five questions on Kepler's laws, g variation, potential and satellites. Attempt them before rereading the formula sheet.