Online Test: Gravitation | Class 11 Physics

Question 1 of 20

The universal gravitational constant $G$ has the value:

$9.8\ \text{N/kg}$

$6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2}$

$6.67\times10^{11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2}$

$1.6\times10^{-19}\ \text{C}$

Explanation: $G=6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2}$ everywhere in the universe.

Question 2 of 20

If the distance between two masses is halved, the gravitational force becomes:

half

double

one-fourth

four times

Explanation: $F\propto\frac{1}{r^2}$; halving $r$ multiplies $F$ by $4$.

Question 3 of 20

The acceleration due to gravity at the surface of the Earth is:

$g=\frac{GM}{R}$

$g=\frac{GM}{R^2}$

$g=\frac{GM^2}{R}$

$g=\frac{GMm}{R^2}$

Explanation: From $mg=\frac{GMm}{R^2}$, $g=\frac{GM}{R^2}$.

Question 4 of 20

The value of $g$ at a height $h$ ($h\ll R$) is approximately:

$g\left(1+\frac{2h}{R}\right)$

$g\left(1-\frac{2h}{R}\right)$

$g\left(1-\frac{h}{R}\right)$

$g\left(1-\frac{h}{2R}\right)$

Explanation: $g_h=g\left(1-\frac{2h}{R}\right)$ for small heights.

Question 5 of 20

At the centre of the Earth the value of $g$ is:

maximum

equal to surface value

zero

infinite

Explanation: $g_d=g\left(1-\frac{d}{R}\right)$ becomes zero at $d=R$, the centre.

Question 6 of 20

The value of $g$ is greatest at the:

equator

poles

Tropic of Capricorn

a mountain top

Explanation: Rotation lowers effective $g$ by $R\omega^2\cos^2\lambda$, which is zero at the poles.

Question 7 of 20

A body of mass 50 kg has a weight on the Moon ($g_{moon}=1.6\ \text{m/s}^2$) of:

$50\ \text{N}$

$80\ \text{N}$

$490\ \text{N}$

$8\ \text{N}$

Explanation: Weight $=mg=50\times1.6=80\ \text{N}$.

Question 8 of 20

The gravitational potential energy of two masses at separation $r$ is:

$\frac{GMm}{r}$

$-\frac{GMm}{r}$

$-\frac{GMm}{r^2}$

$mgr$

Explanation: $U=-\frac{GMm}{r}$, taking PE zero at infinity.

Question 9 of 20

Gravitational potential $V$ at distance $r$ from mass $M$ is:

$-\frac{GM}{r}$

$\frac{GM}{r}$

$-\frac{GM}{r^2}$

$-\frac{GMm}{r}$

Explanation: $V=-\frac{GM}{r}$, measured in J/kg.

Question 10 of 20

Escape velocity from the Earth's surface is given by:

$\sqrt{gR}$

$\sqrt{2gR}$

$\sqrt{\frac{gR}{2}}$

$2gR$

Explanation: $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$.

Question 11 of 20

The escape velocity from the Earth is approximately:

$7.9\ \text{km/s}$

$9.8\ \text{km/s}$

$22.4\ \text{km/s}$

$11.2\ \text{km/s}$

Explanation: $v_e\approx11.2\ \text{km/s}$ for the Earth.

Question 12 of 20

Escape velocity depends on:

the mass of the body

the direction of projection

the mass and radius of the planet

the body's initial height only

Explanation: $v_e=\sqrt{2gR}$ depends only on the planet, not the projectile.

Question 13 of 20

The orbital velocity of a satellite at radius $r$ is:

$\sqrt{\frac{2GM}{r}}$

$\sqrt{\frac{GM}{r}}$

$\frac{GM}{r}$

$\sqrt{GMr}$

Explanation: $v_o=\sqrt{\frac{GM}{r}}$ from gravity providing the centripetal force.

Question 14 of 20

For a satellite very close to the Earth's surface, the orbital velocity is about:

$3\ \text{km/s}$

$5.6\ \text{km/s}$

$11.2\ \text{km/s}$

$7.9\ \text{km/s}$

Explanation: $v_o=\sqrt{gR}\approx7.9\ \text{km/s}$.

Question 15 of 20

The total energy of a satellite in a stable orbit is:

$+\frac{GMm}{2r}$

$-\frac{GMm}{2r}$

$-\frac{GMm}{r}$

zero

Explanation: $E=KE+PE=-\frac{GMm}{2r}$, which is negative (bound).

Question 16 of 20

Kepler's second law (equal areas in equal times) is a result of conservation of:

energy

linear momentum

angular momentum

mass

Explanation: A central force exerts no torque, so angular momentum is conserved.

Question 17 of 20

Kepler's third law states that:

$T^2\propto r^2$

$T^2\propto r^3$

$T^3\propto r^2$

$T\propto r^3$

Explanation: $T^2\propto r^3$, so $\frac{T^2}{r^3}$ is constant.

Question 18 of 20

A geostationary satellite orbits with a period of:

12 hours

24 hours

1 hour

27.3 days

Explanation: Its 24-hour period matches Earth's rotation so it appears stationary.

Question 19 of 20

If a planet's orbital radius around the Sun is 4 times Earth's, its period is:

4 years

8 years

16 years

64 years

Explanation: $T=r^{3/2}=4^{3/2}=8$ years by Kepler's third law.

Question 20 of 20

The relation between escape velocity and orbital velocity near the surface is:

$v_e=v_o$

$v_e=2v_o$

$v_e=\sqrt{2}\,v_o$

$v_e=\frac{v_o}{2}$

Explanation: $v_e=\sqrt{2gR}$ and $v_o=\sqrt{gR}$, so $v_e=\sqrt{2}\,v_o$.

Online Test — Gravitation