*Question*

The Sun has a radius of 7.0×10^{8}m and is a distance 1.5×10^{11} m from Earth. The surface temperature of the Sun is 5800 K.

a.

Show that the intensity of the solar radiation incident on the upper atmosphere of the Earth is approximately 1400Wm^{−2}.[2]

The albedo of the atmosphere is 0.30. Deduce that the average intensity over the entire surface of the Earth is 245Wm^{−2}.[2]

Estimate the average surface temperature of the Earth.[2]

**Answer/Explanation**

### Markscheme

a.

\(I = \frac{{\sigma A{T^4}}}{{4\pi {d^2}}}\)

\( = \frac{{5.67 \times {{10}^{ – 8}} \times {{\left( {7.0 \times {{10}^8}} \right)}^2} \times {{5800}^4}}}{{{{\left( {1.5 \times {{10}^{11}}} \right)}^2}}}\)

* OR *\( \frac{{5.67 \times {{10}^{ – 8}} \times 4\pi \times {{\left( {7.0 \times {{10}^8}} \right)}^2} \times {{5800}^4}}}{{4\pi \times {{\left( {1.5 \times {{10}^{11}}} \right)}^2}}}\)

*I*=1397 Wm^{−2}

*In this question we must see 4SF to award MP3.**Allow candidate to add radius of Sun to Earth–Sun distance. Yields 1386 Wm ^{–2.}*

«transmitted intensity =» 0.70 × 1400 «= 980Wm^{–2}»

\(\frac{{\pi {R^2}}}{{4\pi {R^2}}} \times 980{\rm{W}}{{\rm{m}}^{ – 2}}\)

245Wm^{–2}

5.67 × 10^{–8} × *T*^{4} = 245

*T* = 256K

*Question*

This question is in **two** parts. **Part 1** is about solar radiation and the greenhouse effect. **Part 2** is about a mass on a spring.

**Part 1** Solar radiation and the greenhouse effect

The following data are available.

**Part 2** A mass on a spring

An object is placed on a frictionless surface and attached to a light horizontal spring.

The other end of the spring is attached to a stationary point P. Air resistance is negligible. The equilibrium position is at O. The object is moved to position Y and released.

a.

State the Stefan-Boltzmann law for a black body.[2]

Deduce that the solar power incident per unit area at distance *d* from the Sun is given by

\[\frac{{\sigma {R^2}{T^4}}}{{{d^2}}}\][2]

Calculate, using the data given, the solar power incident per unit area at distance *d* from the Sun.[2]

State **two** reasons why the solar power incident per unit area at a point on the surface of the Earth is likely to be different from your answer in (c).[2]

The average power absorbed per unit area at the Earth’s surface is 240Wm^{–2}. By treating the Earth’s surface as a black body, show that the average surface temperature of the Earth is approximately 250K.[2]

Explain why the actual surface temperature of the Earth is greater than the value in (e).[3]

Outline the conditions necessary for the object to execute simple harmonic motion.[2]

The sketch graph below shows how the displacement of the object from point O varies with time over three time periods.

(i) Label with the letter A a point at which the magnitude of the acceleration of the object is a maximum.

(ii) Label with the letter V a point at which the speed of the object is a maximum.

(iii) Sketch on the same axes a graph of how the displacement varies with time if a **small** frictional force acts on the object.[4]

Point P now begins to move from side to side with a small amplitude and at a variable driving frequency *f*. The frictional force is still small.

At each value of *f*, the object eventually reaches a constant amplitude *A*.

The graph shows the variation with *f* of *A*.

(i) With reference to resonance and resonant frequency, comment on the shape of the graph.

(ii) On the same axes, draw a graph to show the variation with *f* of *A* when the frictional force acting on the object is increased.[4]

**Answer/Explanation**

### Markscheme

a.

power/energy per second emitted proportional to surface area;

and proportional to fourth power of absolute temperature / temperature in K;

Accept equation with symbols defined.

solar power given by 4πR2σT 4 ;

spreads out over sphere of surface area 4πd 2 ;

Hence equation given.

\(\left( {\frac{{\sigma {R^2}{T^4}}}{{{d^2}}} = } \right)\frac{{5.7 \times {{10}^{ – 8}} \times {{\left[ {7.0 \times {{10}^8}} \right]}^2} \times {{\left[ {5.8 \times {{10}^3}} \right]}^4}}}{{{{\left[ {1.5 \times {{10}^{11}}} \right]}^2}}}\);

=1.4×10^{3}(Wm^{-2 });

Award [2] for a bald correct answer.

some energy reflected;

some energy absorbed/scattered by atmosphere; depends on latitude;

depends on time of day;

depends on time of year;

depends on weather (*eg* cloud cover) at location; power output of Sun varies;

Earth-Sun distance varies;

power radiated = power absorbed;

\(T = {}^4\sqrt {\frac{{240}}{{5.7 \times {{10}^{ – 8}}}}} = \left( {250{\rm{K}}} \right)\);

Accept answers given as 260 (K).

radiation from Sun is re-emitted from Earth at longer wavelengths; greenhouse gases in the atmosphere absorb some of this energy; and radiate some of it back to the surface of the Earth;

the force (of the spring on the object)/acceleration (of the object/point O) must be proportional to the displacement (from the equilibrium position/centre/point O);

and in the opposite direction to the displacement / always directed towards the equilibrium position/centre/point O;

(i) one A correctly shown;

(ii) one V correctly shown;

(iii) same period; *(judge by eye)*

amplitude decreasing with time;

(i) resonance is where driving frequency equals/is close to natural/resonant frequency;

the natural/resonant frequency is at/near the maximum amplitude of the graph;

(ii) lower amplitude everywhere on graph, bit still positive;

maximum in same place/moved slightly *(that is, between the lines)* to left on graph;