Questions & explanations
1. Which of the following correctly defines the gravitational potential at a point in a gravitational field?
- The work done by gravitational force in bringing a unit mass from infinity to that point.
- The work done against gravitational force in bringing a unit mass from infinity to that point.
- The gravitational force per unit mass acting on a test mass at that point.
- The negative of the gravitational field intensity at that point.
Answer: The work done against gravitational force in bringing a unit mass from infinity to that point.
Gravitational potential is defined as the work done by an external agent against the gravitational force in bringing a unit test mass from infinity to a given point without acceleration. It is a scalar quantity, measured in J/kg. Option A gives the work done by gravitational force, which is the negative of the potential. Option C defines gravitational field intensity. Option D is not a standard definition; field intensity is the negative gradient of potential, not the negative of potential itself.
2. Two point masses of 5 kg each are placed 2 m apart in free space. What is the gravitational potential energy of the system? (Take G = 6.67 × 10^{-11} N m^2 kg^{-2})
- -8.34 × 10^{-10} J
- -6.67 × 10^{-10} J
- -4.17 × 10^{-10} J
- -2.00 × 10^{-10} J
Answer: -8.34 × 10^{-10} J
The gravitational potential energy of a two-body system is given by U = -G m1 m2 / r. Substituting values: m1 = m2 = 5 kg, r = 2 m. U = - (6.67 × 10^{-11} × 5 × 5) / 2 = - (6.67 × 10^{-11} × 25) / 2 = - (166.75 × 10^{-11}) / 2 = -83.375 × 10^{-11} J = -8.3375 × 10^{-10} J, which rounds to -8.34 × 10^{-10} J. Options B, C, D have incorrect numerical values.
3. What is the escape velocity from the surface of Earth? (Given: G = 6.67 × 10⁻¹¹ N m²/kg², M = 5.98 × 10²⁴ kg, R = 6.4 × 10⁶ m)
- 7.9 km/s
- 11.2 km/s
- 8.9 km/s
- 15.0 km/s
Answer: 11.2 km/s
Escape velocity vₑ = √(2GM/R). Substituting the given values: vₑ = √(2 × 6.67×10⁻¹¹ × 5.98×10²⁴ / 6.4×10⁶) ≈ 1.12×10⁴ m/s = 11.2 km/s. Option A is the orbital velocity for a near‑Earth satellite, Option C (8.9 km/s) is simply a plausible-looking distractor value; it is not the Moon's escape velocity (which is about 2.4 km/s)., and option D is too high.
4. The binding energy of a satellite of mass m in a circular orbit of radius r around Earth (mass M) is:
- GMm/r
- GMm/(2r)
- -GMm/(2r)
- 2GMm/r
Answer: GMm/(2r)
Binding energy is the minimum energy required to move the satellite from its orbit to infinity, which equals the negative of the total mechanical energy. Total energy E = -GMm/(2r), so binding energy = -E = GMm/(2r). Options A and D are multiples of this, and option C is negative (energy cannot be negative as a positive amount is required).
5. The acceleration due to gravity measured at a latitude φ on the Earth's surface, considering the Earth's rotation with angular speed ω, is given by:
- g = g₀ - R ω² cos² φ
- g = g₀ - R ω² sin² φ
- g = g₀ - R ω² cos φ
- g = g₀ - R ω² sin φ
Answer: g = g₀ - R ω² cos² φ
The centrifugal reduction depends on the component of the Earth's rotation perpendicular to the local vertical. The effective g is g₀ - R ω² cos² φ. At the equator (φ = 0), the reduction is maximum; at poles (φ = 90°), the reduction is zero. Option B gives reduction maximum at poles, which is wrong; C and D have wrong trig functions.
6. For a satellite orbiting Earth in a circular orbit at a height h above the surface, the orbital velocity v₀ is given by (where R = Earth's radius, M = Earth's mass, G = gravitational constant, g = acceleration due to gravity at the surface):
- v₀ = √(GM/(R+h))
- v₀ = √(2GM/(R+h))
- v₀ = √(g(R+h))
- v₀ = √(2gR)
Answer: v₀ = √(GM/(R+h))
For a circular orbit, the centripetal force required (mv₀²/(R+h)) is provided by the gravitational force (GMm/(R+h)²). Equating and solving gives v₀ = √(GM/(R+h)). Option B is the escape velocity formula, option C is dimensionally incorrect (g(R+h) gives m²/s²), and option D is the near‑surface orbital velocity ignoring height.
7. A person inside an orbiting satellite feels weightless because:
- The satellite is far from Earth's gravity
- The gravitational force on the person is zero
- The person and the satellite are in free fall towards Earth with the same acceleration
- The person's mass becomes zero
Answer: The person and the satellite are in free fall towards Earth with the same acceleration
Weightlessness occurs because both the satellite and the person inside it are falling freely towards Earth under gravity, with the same acceleration. Hence there is no normal reaction force from the floor, giving the sensation of weightlessness. Gravitational force is still present; neither mass nor distance are zero.
8. Which of the following correctly differentiates acceleration due to gravity (g) from the universal gravitational constant (G)?
- g is constant everywhere, while G varies with location.
- g is a scalar quantity, while G is a vector constant.
- g depends on the mass of the falling object, while G is independent of mass.
- g has dimensions [M⁰ L¹ T⁻²], while G has dimensions [M⁻¹ L³ T⁻²].
Answer: g has dimensions [M⁰ L¹ T⁻²], while G has dimensions [M⁻¹ L³ T⁻²].
g has dimensions of acceleration (LT⁻²) and G has dimensions derived from F r²/m², which is M⁻¹ L³ T⁻². Option A is false because g varies with location; B is false because g is a vector and G is a scalar; C is false because g is independent of the object's mass (free-fall acceleration is the same for all objects).
9. The value and SI unit of the universal gravitational constant G is:
- 6.67 × 10⁻¹¹ N m² kg⁻²
- 6.67 × 10⁻¹¹ N m kg⁻²
- 6.67 × 10⁻¹¹ N m² kg⁻¹
- 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²
Answer: 6.67 × 10⁻¹¹ N m² kg⁻²
From F = G m1 m2 / r², G = F r² / (m1 m2). The SI unit of force is N, distance is m, mass is kg, so unit of G is N m² kg⁻². Numerically, G = 6.67 × 10⁻¹¹ in these units. Option B is missing one metre, C has kg⁻¹ instead of kg⁻², and D has m² instead of m³ and is dimensionally inconsistent.
10. Which of the following correctly represents Newton's universal law of gravitation in vector form? (Take vector F as the gravitational force on mass m2 due to mass m1, and vector r as the position vector of m2 relative to m1.)
- F = (G m1 m2 / r^2) r̂
- F = - (G m1 m2 / r^2) r̂
- F = (G m1 m2 / r^3) r⃗
- F = - (G m1 m2 / r) r̂
Answer: F = - (G m1 m2 / r^2) r̂
The correct vector form includes a negative sign to indicate the attractive nature and uses the unit vector r̂. Option B matches this exactly. Option A omits the negative sign, C uses the vector form without the negative sign, and D has an incorrect power of r in the denominator.
11. Gravitational field intensity at a point is defined as:
- The gravitational force per unit mass acting on a test mass placed at that point.
- The acceleration due to gravity at that point.
- Both A and B.
- The gravitational potential energy per unit mass at that point.
Answer: Both A and B.
Gravitational field intensity is defined as the force per unit mass, and its magnitude equals the acceleration due to gravity at that point. Thus both A and B are correct definitions; C is the correct choice. Option D defines gravitational potential, not field intensity.
12. A satellite is orbiting Earth at a height h above its surface. If R is the radius of Earth and T is the orbital period of the satellite, which of the following correctly gives the orbital speed v of the satellite?
- v = 2πR / T
- v = 2π(R + h) / T
- v = 2π(R + h) / T²
- v = 2πT / (R + h)
Answer: v = 2π(R + h) / T
The satellite moves in a circular orbit of radius (R + h). Its orbital speed is the circumference of the orbit divided by the time period: v = 2π(R + h) / T. The other options incorrectly use Earth's radius alone, include T², or invert the relationship.