# IBDP Physics 6.2 – Newton’s law of gravitation: IB Style Question Bank HL Paper 1

### Question

The centre of the Earth is separated from the centre of the Moon by a distance D. Point P lies on a line joining the centre of the Earth and the centre of the Moon, a distance X from the centre of the Earth. The gravitational field strength at P is zero.

What is the ratio $$\frac{{{\text{mass of the Moon}}}}{{{\text{mass of the Earth}}}}$$?

A.  $$\frac{{{{\left( {D – X} \right)}^2}}}{{{X^2}}}$$

B.  $$\frac{{\left( {D – X} \right)}}{X}$$

C.  $$\frac{{{X^2}}}{{{{\left( {D – X} \right)}^2}}}$$

D.  $$\frac{X}{{D – X}}$$

### Markscheme

A

Since The gravitational field strength at P is zero Hence

$$\frac{GM_e}{x^2}=\frac{GM_m}{({D-x})^{2}}$$
$$\therefore \frac{M_m}{M_e}=\frac{({D-x})^{2}}{x^2}$$

### Question

Which single condition enables Newton’s universal law of gravitation to be used to predict the force between the Earth and the Sun?

A. The Earth and the Sun both have a very large radius.

B. The distance between the Earth and the Sun is approximately constant.

C. The Earth and the Sun both have a very large mass.

D. The Earth and the Sun behave as point masses.

### Markscheme

D

Although Newton’s law of gravitation applies strictly to particles, we can also apply it to real objects as long as the sizes of the objects are small relative to the distance between them.The Sun and Earth are far enough apart so that, to a good approximation, we can treat them both as particles

### Question

The Earth is a distance $${r_S}$$ from the Sun. The Moon is a distance $${r_M}$$ from the Earth.

The ratio $$\frac{{{\text{gravitational field strength at the Earth due to the Sun}}}}{{{\text{gravitational field strength at the Earth due to the Moon}}}}$$ is proportional to

A.     $$\frac{{{r_M}}}{{{r_S}}}$$

B.     $$\frac{{{r_S}}}{{{r_M}}}$$

C.     $$\frac{{r_{\text{S}}^2}}{{r_{^M}^2}}$$.

D.     $$\frac{{r_{\text{M}}^2}}{{r_{^S}^2}}$$

### Markscheme

D

gravitational field strength at the Earth due to the Sun = $$\frac{GM_s}{r_{s}^2}$$

gravitational field strength at the Earth due to the Moon  = $$\frac{GM_m}{r_{M}^2}$$

Hence ratio is

$$\frac{gravitational field strength at the Earth due to the Sun}{gravitational field strength at the Earth due to the Moon}$$

$$=\frac{\frac{GM_S}{r_{S}^2}}{\frac{GM_m}{r_{M}^2}}$$

$$=\frac{r_{M}^2}{r_{S}^2}$$

### Question

The acceleration of free fall of a mass of 2.0 kg close to the surface of Mars is 3.6 ms–2. What is the gravitational field strength at the surface of Mars in N kg–1?

A. 1.8
B. 3.6
C. 7.2
D. 9.8

### Markscheme

B

Gravitational field intensity of the field is given by

$$\frac{\vec F}{m} =\frac{m\vec g}{m}= \vec g$$

Thus, the intensity of the gravitational field near the surface of the earth/any planet is equal to the acceleration due to gravity. It should be clearly understood that the intensity of the gravitational field and the acceleration due to gravity are two separate physical quantities having equal magnitudes and directions

Hence as given in questions $$g=3.6$$ , hence Field Intensity $$= g = 3.6$$

### Question

An astronaut of mass 60 kg is on board the International Space Station, which is in low orbit around the Earth. The gravitational force of attraction between the Earth and astronaut is approximately

A. zero.
B. 6 N.
C. 60 N.
D. 600 N.

gravitational force of attraction is given by $$F=\frac{GM_eM_a}{R_e^2}$$ R is taken as Radius of Earth since astronaut is in lower orbit.
Now this is equal to $$mg$$
On surface of earth (in in lower orbit g is taken as $$10 \; Nm^{-2}$$
$$= 60 \times 10 = 600 N$$