(a) Two point masses are isolated in space and are separated by a distance $x$. State an expression relating the gravitational force $F$ between the two masses to the magnitudes $M$ and $m$ of the masses. State the name of any other symbol used. (b) A spacecraft is to be put into a circular orbit about a spherical planet. The planet may be considered to be isolated in space. The mass of the planet, assumed to be concentrated at its centre, is $7.5 \times 10^{23}\ \text{kg}$.…

(a) F = GMm/x^2; G is the universal gravitational constant. · (b)(i) 3.7 × 10^3 m s⁻1 · (b)(ii) ΔGPE = −8.3 × 10^9 J (decrease of 8.3 × 10^9 J) · (c) Decreases.

Physics 9702/43 — May/June 2019

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

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Topics Magnetic Fields · Gravitational Fields · Motion in a Circle · Ideal Gases · Thermodynamics · Oscillations · +7 more

Q1Gravitational FieldsMotion in a CircleFree sample

(a)

Two point masses are isolated in space and are separated by a distance $x$ .

State an expression relating the gravitational force $F$ between the two masses to the magnitudes $M$ and $m$ of the masses. State the name of any other symbol used.

DifficultyEasy

Worked solution

Answer

F = \frac{GMm}{x^2}

where $G$ is the universal gravitational constant.

Final answer

F = GMm/x^2; G is the universal gravitational constant.

Detailed explanation

Background Concept

Newton's law of gravitation gives the magnitude of the force between two point masses separated by a distance $x$ :

F = \frac{Gm_1 m_2}{x^2}

$F$ is the magnitude of the gravitational force (always attractive).
$m_1$ and $m_2$ are the masses.
$x$ is the separation between their centres.
$G$ is the universal gravitational constant.

Understanding the Question

You are told there are two isolated point masses, of magnitudes $M$ and $m$ , separated by a distance $x$ . You must state the relationship for the gravitational force between them and name any extra symbol used.

Approach

Use the inverse-square law for gravitation and replace $m_1, m_2$ with $M, m$ . Then identify $G$ .

Step-by-Step Reasoning

Start from Newton's law:

F = \frac{Gm_1 m_2}{x^2}

Substitute $m_1 = M$ and $m_2 = m$ :

F = \frac{GMm}{x^2}

The additional symbol is $G$ , called the universal gravitational constant.

Key Takeaways

Gravitational force between point masses follows an inverse-square dependence on separation.
Always name physical constants when asked.

Common Mistakes

Writing $F = GMm/x$ (missing the square).
Forgetting to mention what $G$ represents.

Things to Be Careful About

$x$ is the distance between the centres of the masses.
The question asks for an expression for the force magnitude, so no direction is needed here.

Techniques used

apply Newton's law of gravitationidentify and name physical constants in an equation

(b)

A spacecraft is to be put into a circular orbit about a spherical planet.

The planet may be considered to be isolated in space. The mass of the planet, assumed to be concentrated at its centre, is $7.5 \times 10^{23}\ \text{kg}$ . The radius of the planet is $3.4 \times 10^6\ \text{m}$ .

(i)

The spacecraft is to orbit the planet at a height of $2.4 \times 10^5\ \text{m}$ above the surface of the planet. At this altitude, there is no atmosphere.

Show that the speed of the spacecraft in its orbit is $3.7 \times 10^3\ \text{m s}^{-1}$ .

DifficultyMedium-Easy

Worked solution

Working

Orbital radius

r = 3.4 \times 10^6 + 2.4 \times 10^5 = 3.64 \times 10^6\ \text{m}

For a circular orbit,

\frac{GMm}{r^2} = \frac{mv^2}{r}

v = \sqrt{\frac{GM}{r}}

v = \sqrt{\frac{(6.67 \times 10^{-11})(7.5 \times 10^{23})}{3.64 \times 10^6}} = 3.7 \times 10^3\ \text{m s}^{-1}

Answer

$3.7 \times 10^3\ \text{m s}^{-1}$

Final answer

3.7 × 10^3 m s⁻1

Detailed explanation

Background Concept

For a spacecraft in a circular orbit, the inward (centripetal) force needed is provided by gravity.

Gravitational force on mass $m$ at distance $r$ from planet centre:

F_g = \frac{GMm}{r^2}

Centripetal force needed for speed $v$ in a circle of radius $r$ :

F_c = \frac{mv^2}{r}

Setting $F_g = F_c$ gives the standard circular-orbit speed:

v = \sqrt{\frac{GM}{r}}

Understanding the Question

The planet has mass $M = 7.5 \times 10^{23}\ \text{kg}$ and radius $R = 3.4 \times 10^6\ \text{m}$ . The spacecraft orbits at height $h = 2.4 \times 10^5\ \text{m}$ above the surface, so the orbital radius from the centre is $r = R + h$ . You must show the orbital speed is $3.7 \times 10^3\ \text{m s}^{-1}$ .

Approach

Calculate orbital radius $r = R + h$ .
Use $F_g = F_c$ to derive $v = \sqrt{GM/r}$ .
Substitute values (including $G$ ) and evaluate.

Step-by-Step Reasoning

Radius from centre:

r = 3.4 \times 10^6 + 2.4 \times 10^5 = 3.64 \times 10^6\ \text{m}

Equate forces:

\frac{GMm}{r^2} = \frac{mv^2}{r}

Cancel $m$ (spacecraft mass does not affect orbital speed) and rearrange:

\frac{GM}{r} = v^2 \quad \Rightarrow \quad v = \sqrt{\frac{GM}{r}}

Substitute numbers:

GM = (6.67 \times 10^{-11})(7.5 \times 10^{23}) = 5.00 \times 10^{13}

\frac{GM}{r} = \frac{5.00 \times 10^{13}}{3.64 \times 10^6} \approx 1.37 \times 10^7

v = \sqrt{1.37 \times 10^7} \approx 3.7 \times 10^3\ \text{m s}^{-1}

Key Takeaways

For circular orbit: gravity provides centripetal force.
Orbital speed depends on $M$ and $r$ only: $v = \sqrt{GM/r}$ .

Common Mistakes

Using $r = 2.4 \times 10^5\ \text{m}$ instead of $r = R + h$ .
Forgetting to cancel the spacecraft mass $m$ .
Using $r^2$ in the final expression by algebra error.

Things to Be Careful About

Always measure $r$ from the planet's centre.
Check powers of ten when multiplying $G$ and $M$ .
Final speed should be to 2 s.f. or consistent with the given value $3.7 \times 10^3\ \text{m s}^{-1}$ .

Techniques used

equate gravitational force to centripetal force for a circular orbitsubstitute orbital radius as planet radius plus altituderearrange to obtain orbital speed and evaluate numerically

(ii)

One possible path of the spacecraft as it approaches the planet is shown in Fig. 1.1.

Fig. 1.1

The spacecraft enters the orbit at point A with speed $3.7 \times 10^3\ \text{m s}^{-1}$ .

At point B, a distance of $5.00 \times 10^7\ \text{m}$ from the centre of the planet, the spacecraft has a speed of $4.1 \times 10^3\ \text{m s}^{-1}$ . The mass of the spacecraft is $650\ \text{kg}$ .

For the spacecraft moving from point B to point A, show that the change in gravitational potential energy of the spacecraft is $8.3 \times 10^9\ \text{J}$ .

DifficultyMedium

Worked solution

Working

U = -\frac{GMm}{r}

$r_A = 3.64 \times 10^6\ \text{m}$ , $r_B = 5.00 \times 10^7\ \text{m}$ .

Change in GPE from $B$ to $A$ :

\Delta U = U_A - U_B = -GMm\left(\frac{1}{r_A}-\frac{1}{r_B}\right)

\Delta U = -(6.67\times 10^{-11})(7.5\times 10^{23})(650) \left(\frac{1}{3.64\times 10^6}-\frac{1}{5.00\times 10^7}\right)

\Delta U \approx -8.3 \times 10^9\ \text{J}

Answer

Gravitational potential energy decreases by $8.3 \times 10^9\ \text{J}$ (i.e. $\Delta U = -8.3 \times 10^9\ \text{J}$ ).

Final answer

ΔGPE = −8.3 × 10^9 J (decrease of 8.3 × 10^9 J)

Detailed explanation

Background Concept

The gravitational potential energy (GPE) of a mass $m$ in the gravitational field of a spherical mass $M$ (with $m$ outside the sphere) is defined relative to zero at infinity:

U = -\frac{GMm}{r}

where $r$ is the distance from the centre of the planet.

A change in GPE when moving from radius $r_1$ to $r_2$ is:

\Delta U = U_2 - U_1 = -GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right)

As you get closer to the planet ( $r$ decreases), $U$ becomes more negative, meaning GPE decreases.

Understanding the Question

Point $B$ is far from the planet centre: $r_B = 5.00 \times 10^7\ \text{m}$ . Point $A$ is the circular orbit radius found earlier: $r_A = 3.64 \times 10^6\ \text{m}$ . The spacecraft mass is $m = 650\ \text{kg}$ . You must show that, moving from $B$ to $A$ , the change in GPE has magnitude $8.3 \times 10^9\ \text{J}$ .

Approach

Use $U = -GMm/r$ at the two positions.
Compute $\Delta U = U_A - U_B$ .
Expect a negative result (because the spacecraft moves closer), and report the magnitude consistent with the required value.

Step-by-Step Reasoning

Write the two GPE values:

U_A = -\frac{GMm}{r_A}, \quad U_B = -\frac{GMm}{r_B}

So the change from $B$ to $A$ is

\Delta U = U_A - U_B = -GMm\left(\frac{1}{r_A} - \frac{1}{r_B}\right)

Substitute values ( $G = 6.67\times 10^{-11}$ , $M = 7.5\times 10^{23}$ , $m = 650$ ):

GMm = (6.67\times 10^{-11})(7.5\times 10^{23})(650) \approx 3.25 \times 10^{16}

Compute the bracket:

\frac{1}{r_A} - \frac{1}{r_B} = \frac{1}{3.64\times 10^6} - \frac{1}{5.00\times 10^7} \approx 2.75\times 10^{-7} - 2.00\times 10^{-8} = 2.55\times 10^{-7}

Then

\Delta U \approx -(3.25\times 10^{16})(2.55\times 10^{-7}) \approx -8.3\times 10^9\ \text{J}

The negative sign means GPE decreases; the magnitude of the change is $8.3\times 10^9\ \text{J}$ .

Key Takeaways

Use $U = -GMm/r$ with $r$ measured from the centre.
Moving closer to the planet makes $U$ more negative (a decrease in GPE).

Common Mistakes

Using $r_A$ as height above the surface rather than distance from the centre.
Dropping the minus sign and claiming GPE increases.
Using $\Delta U = GMm(1/r_A - 1/r_B)$ without checking sign convention.

Things to Be Careful About

The question wording may quote the magnitude; always state whether it is an increase or decrease.
Keep enough significant figures during intermediate steps to obtain $8.3 \times 10^9\ \text{J}$ to 2 s.f.

Techniques used

use gravitational potential energy U = -GMm/rcalculate change in potential energy between two radiievaluate and interpret the sign of an energy change

(c)

By considering changes in gravitational potential energy and in kinetic energy of the spacecraft, determine whether the total energy of the spacecraft increases or decreases in moving from point B to point A. A numerical answer is not required.

DifficultyMedium-Easy

Worked solution

Answer

From $B$ to $A$ , the spacecraft moves closer to the planet so its gravitational potential energy decreases (becomes more negative).

Also $v$ decreases from $4.1 \times 10^3\ \text{m s}^{-1}$ to $3.7 \times 10^3\ \text{m s}^{-1}$ , so kinetic energy decreases.

Therefore the total energy ( $\text{KE} + \text{GPE}$ ) decreases.

Final answer

Decreases.

Detailed explanation

Background Concept

The total mechanical energy of the spacecraft in the planet's gravitational field is

E = K + U

where

K = \frac{1}{2}mv^2, \quad U = -\frac{GMm}{r}

If speed decreases, kinetic energy decreases.
If the spacecraft moves closer to the planet ( $r$ decreases), $U$ becomes more negative, so GPE decreases.

Understanding the Question

You are told the spacecraft moves from point $B$ (far away, $r_B = 5.00\times 10^7\ \text{m}$ , speed $4.1\times 10^3\ \text{m s}^{-1}$ ) to point $A$ (orbit radius, speed $3.7\times 10^3\ \text{m s}^{-1}$ ). You must decide whether the spacecraft's total energy increases or decreases, without calculating a numerical value.

Approach

Determine the sign of the change in GPE from $B$ to $A$ (closer means GPE decreases).
Compare speeds to decide the sign of the change in KE.
Combine: if both $K$ and $U$ decrease, their sum decreases.

Step-by-Step Reasoning

Gravitational potential energy:
Moving from $B$ to $A$ reduces $r$ , so

U = -\frac{GMm}{r}

becomes more negative. Therefore $\Delta U = U_A - U_B$ is negative: GPE decreases.

Kinetic energy:
Speed changes from $4.1\times 10^3$ to $3.7\times 10^3\ \text{m s}^{-1}$ . Since

K = \frac{1}{2}mv^2,

a smaller $v$ means smaller $K$ . Therefore kinetic energy decreases.

Total energy:
Since both components decrease,

\Delta E = \Delta K + \Delta U

is negative, so the total energy decreases.

Key Takeaways

Use signs: closer to a gravitating mass means GPE decreases.
Comparing speeds is enough to decide the direction of KE change.

Common Mistakes

Thinking GPE increases when the spacecraft falls inward (it actually decreases because it becomes more negative).
Assuming KE must increase when moving closer; here the given speeds show the opposite.

Things to Be Careful About

The question asks about the spacecraft's total energy, not just the magnitude of energy change.
Always use the given speeds rather than assuming what “should” happen based on circular orbit ideas.

Techniques used

compare kinetic energies using speeds at two pointsuse the sign of gravitational potential energy changededuce the change in total mechanical energy from energy component changes

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