DC Pandey Solutions: Gravitation - 1 | Physics Class 11 - NEET PDF Download

Introductory Exercise 10.1

Q.1. Calculate the change in the value of g at altitude 45°. Take radius of earth = 6.37 × 10³ km

Sol. When the Earth rotates, the effective acceleration due to gravity at latitude φ is reduced by the centrifugal term ω²Rcos²φ. The change in g between the equator (φ = 0°) and latitude φ is therefore Δg = ω²R(1 - cos²φ) = ω²R sin²φ.
Using ω = 7.2921 × 10^-5 s^-1 and R = 6.37 × 10⁶ m:
ω²R ≈ 0.03386 m/s².
For φ = 45°, sin²45° = 1/2, so
Δg = 0.03386 × 1/2 ≈ 0.0169 m/s².
Thus the change in the value of g is about 0.0168 m/s².

Q.2. Determine the speed with which the earth would have to rotate on its axis, so that a person on the equator would weigh 3/5th as much as at present. Take R = 6400 km

Sol. Let ω be the required angular speed. Effective acceleration at the equator is g' = g - ω²R. We require g' = (3/5)g.
So g - ω²R = (3/5)g ⇒ ω²R = (2/5)g ⇒ ω = sqrt((2g)/(5R)).
Take g = 9.8 m/s², R = 6.4 × 10⁶ m:
ω ≈ sqrt(19.6/(3.2 × 10⁷)) ≈ 7.828 × 10^-4 s^-1.
The linear speed at the equator is v = ωR ≈ 7.828 × 10^-4 × 6.4 × 10⁶ ≈ 5.00 × 10³ m/s ≈ 5.00 km/s.
Optionally, the rotation period T = 2π/ω ≈ 2.23 hours.

Q.3. At what height from the surface of earth will the value of g be reduced by 36% from the value at the surface? R = 6400 km

Sol. Let h be the height above the surface where g' = 0.64 g.
We have g' = g (R/(R+h))² = 0.64 g ⇒ (R/(R+h)) = √0.64 = 0.8.
Hence R + h = R/0.8 = 1.25 R ⇒ h = 0.25 R = R/4.
With R = 6400 km, h = 1600 km.

Q.4. The distance between two bodies A and B is r. Taking the gravitational force according to the law of inverse square of r, the acceleration of the body A is a. If the gravitational force follows an inverse fourth power law, then what will be the acceleration of body A?

Sol: For the inverse-square law, a = k / r², where k is the proportionality constant (k = GM for Newtonian gravity).
If the force instead follows an inverse fourth power with the same proportionality constant k, then the new acceleration is a' = k / r⁴
= (k / r²) × 1/r²
= a / r².
Thus a' = a / r²(assuming the same constant of proportionality is used for both laws).

Q.5. Find the force of attraction on a particle of mass m placed at the centre of a semicircular wire of length L and mass M.

Sol. Let the semicircular wire have radius R and uniform linear mass density λ = M / L. For a semicircle, L = πR, so R = L/π.
Consider an element of the wire of length Rdθ at angle θ (measured from the positive x-axis). Its mass is dm = λR dθ. The magnitude of the gravitational force from dm on the central mass m is dF = G m dm / R² = G m λ dθ / R.
Resolve dF along the symmetry axis (say the y-axis); the component is dF_y = dF sinθ = (G m λ / R) sinθ dθ.
Integrate θ from 0 to π (upper semicircle):
F = ∫₀^π (G m λ / R) sinθ dθ = (G m λ / R) [-cosθ]₀^π = (G m λ / R) (2).
So F = 2 G m λ / R = 2 G m (M / L) / R = 2 G m M /(L R).
Using R = L/π, this becomes F = 2 G m M /(π R²) = 2π G m M / L².
Direction: The resultant force is along the symmetry axis of the semicircle, directed towards the arc.

Introductory Exercise 10.2

Q.1. A particle of mass m is placed at the centre of a uniform spherical shell of same mass and radius R. Find the gravitational potential at a distance R/2 from the centre.

Sol.

V = V_c + V_s=

Explanation:
The potential at a point is the algebraic sum of potentials due to all masses present.
1. Potential due to the central particle (mass m) at distance r = R/2 is V_c = -Gm / (R/2) = -2Gm/R.
2. Potential due to a thin spherical shell of mass equal to m at any point inside the shell is constant and equals -Gm / R.
Thus total potential at r = R/2 is V = -2Gm/R - Gm/R = -3Gm/R.

Q.2. A particle of mass 20 g experiences a gravitational force of 4.0 N along positive x-direction. Find the gravitational field at that point.

Sol. Gravitational field g = F/m. Here m = 20 g = 0.020 kg and F = 4.0 N along +x.
So g = 4.0 / 0.020 = 200 N/kg along +i.
Field = 200 N/kg i.

Q.3. The gravitational potential due to a mass distribution is V = 3x²y + y³z. Find the gravitational field.

Sol. Explanation:
The gravitational field is the negative gradient of the potential: g = -∇V.
Compute partial derivatives:
∂V/∂x = 6xy, ∂V/∂y = 3x² + 3y²z, ∂V/∂z = y³.
Therefore
g = -(6xy) i - (3x² + 3y²z) j - (y³) k.
So the field is g = -6xy i - (3x² + 3y²z) j - y³k(N/kg).

Q.4. Gravitational potential at x = 2 m is decreasing at a rate of 10 J/kg-m along the positive x-direction. It implies that the magnitude of gravitational field at x = 2 m is also 10 N/kg. Is this statement true or false?

Sol. Explanation:
False. The rate of decrease of potential along the positive x-direction equals the x-component of the gravitational field, that is g_x = 10 N/kg. However, the magnitude of the total gravitational field is |g| = sqrt(g_x² + g_y² + g_z²) and could be larger if g_y or g_z are non-zero. Only the x-component is known from the given information, so the statement that the magnitude equals 10 N/kg is not necessarily true.

Q.5. The gravitational potential in a region is given by, V = 20(x+ y) J/kg. Find the magnitude of the gravitational force on a particle of mass 0.5 kg placed at the origin.

Sol.  First find the gravitational field: g = -∇V.

∂V/∂x = 20, ∂V/∂y = 20, ∂V/∂z = 0 ⇒ g = -20 i - 20 j (N/kg).
Magnitude of field = √(20² + 20²) = 20√2 N/kg.
Force on mass m = 0.5 kg is F = m g (vector) = 0.5(-20 i - 20 j) = -10 i - 10 j N.
Magnitude of force = 0.5 × 20√2 = 10√2 N.

Sol. Since the gravitational field is conservative, the work done by the field when a particle of mass m moves from point A to point B is W = m ∫_A^B g · dr = m [V(A) - V(B)], where V is the gravitational potential (with g = -∇V).
Evaluate either the line integral ∫ g · dr along the given straight path or compute the potential at the initial and final points and take their difference multiplied by m. The detailed substitution and algebra are shown in the provided figure .