Gravitation — Class 11 Physics: 17 MCQs with answers

A · It is always attractive and acts along the line joining the masses.

Gravitational force is always attractive and acts along the line joining the two masses; it never repels and needs no contact.

C · $6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$

$G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ is the universal gravitational constant. $9.8\ \text{m s}^{-2}$ is $g$ on Earth and $11.2\ \text{km/s}$ is escape velocity.

B · One-ninth of the original force

Since $F \propto 1/r^2$, tripling $r$ makes the force $1/3^2 = 1/9$ of its original value.

D · Weight is a force measured in newtons; 60 kg is the mass.

Mass is the quantity of matter in kg and is constant; weight is the gravitational force $W = mg$ measured in newtons and changes with $g$.

A · True

The universal gravitational constant $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ is the same everywhere in the universe.

D · $v = \sqrt{\frac{GM}{r}}$

Equating gravitational force to centripetal force, $\frac{GMm}{r^2} = \frac{mv^2}{r}$, gives $v = \sqrt{GM/r}$. The factor of 2 appears only in escape velocity.

B · Its value is the same everywhere in the universe.

$G$ is a fundamental constant with the same value everywhere. In contrast, $g = GM/R^2$ is local and varies with the planet's mass and radius.

B · The astronaut and station are in free fall together, so the normal force is zero.

Apparent weightlessness is free fall, not absence of gravity. Both astronaut and station accelerate toward Earth at the same rate, so no normal force acts.

B · $g = \frac{GM}{R^2}$

Equating $mg$ with $GMm/R^2$ and cancelling $m$ gives $g = GM/R^2$.

D · $U = -\frac{GMm}{r}$

The exact form is $U = -GMm/r$, with the negative sign indicating a bound system. $U = mgh$ is only a small-height approximation.

B · $11.2\ \text{km/s}$

Using $v_e = \sqrt{2GM/R}$ for Earth gives about $11.2\ \text{km/s}$. $2.4\ \text{km/s}$ is the Moon's escape velocity and $3 \times 10^8\ \text{m/s}$ is the speed of light.

B · $T^2 \propto a^3$

Kepler's third law is $T^2 \propto a^3$, which follows from equating gravitational and centripetal forces for an orbit.

A · Gravity is the weakest but dominates on astronomical scales because it is always attractive and has infinite range.

Gravity is the weakest fundamental force, but because it is always attractive and has infinite range, it accumulates over huge masses and governs astronomical structures.

A · Half of Earth's $g$

$g = GM/R^2$. Doubling both $M$ and $R$ gives $g' = G(2M)/(2R)^2 = (2/4)\,GM/R^2 = g/2$.

B · Its tangential velocity keeps it in orbit; gravity acts as the centripetal force.

The Moon continuously falls toward Earth, but its tangential velocity carries it around; gravity provides the centripetal force $GMm/r^2 = mv^2/r$.

B · Both A and R are true, and R correctly explains A.

A black hole's escape velocity $v_e = \sqrt{2GM/R}$ exceeds $c$, so even light cannot escape — R is the correct explanation of A.

A · 7.91 km/s

To find the orbital velocity \( v \) of the satellite, we use the formula \( v = \sqrt{rac{GM}{r}} \). The radius \( r \) is the sum of Earth's radius and the height of the satellite: \( r = 6400 \ ext{km} + 500 \ ext{km} = 6900 \ ext{km} = 6.9 imes 10^6 \ ext{m} \). Plugging in the values, we find \( v = \sqrt{…