Answers

WARNING - the ideas are right, the maths may not be!

1. A planet orbits around a 5 $\rm M_{\odot}$ star at a distance of 5 AU's. What is it's period?

Let anything underscore 1 relate to the Earth-Sun system.

M₂ = 5 $\rm M_{\odot}$ = 5M₁. a₂ = 5AU = 5a₁ so

$$\frac{P^2_1 M_1}{a^3_1} = \frac{P^2_1 5M_1}{(5a_1)^3}$$ (1)

M₁ and a₁ cancel each side leaving P₁² = P₂²5/(5x5x5) = P₂²/25 so P₂ = 5 P₁ = 5 years

2. A planet orbits a star in a Period of 3 years at a distance of 2 AU's. What is the star's mass?

Let anything underscore 1 relate to the Earth-Sun system.

P₂ = 3 years = 3P₁, a₂ = 2AU = 2a₁ so

$$\frac{P^2_1 M_1}{a^3_1} = \frac{(3P_1)^2 M_2}{(2a_1)^3}$$ (2)

P₁ and a₁ cancel so M₁ = 9 M₂/8 so M₂ = 8/9 M₁ = 8/9 $\rm M_{\odot}$.

3. A satellite orbits an 'earth' at a distance of 1000000 km in 30 hours. If we double the distance, what is the new period?

Using the same equations but this time need to set up what subscript 1 and 2 things are. Told in question that satellite orbits at 1000000 km in 30 hours. Therefore a₁ = 1000000 km, P₁ = 30 hours, M₁ unknown. Now second scenario a₂ = 2a₁ (told in question), want to find P₂ and M₂ = M₁ (same earth type object).

so as before

$$\frac{P^2_1 M_1}{a^3_1} = \frac{(P_2)^2 M_1}{(2a_1)^3}$$ (3)

and M₁ and a₁ cancel each side so that P₁ = P₂/8 so P₂ = 8P₁ = 8*30 hours = 240 hours.

4. A satellite orbits a planet at a distance of 1000000 km in 30 hours. If we double the mass of the planet and keep the period the same as in (3), what distance does the satellite orbit at?

Now M₂ = 2M₁ and P₂ = 8P₁ (found from above) find a₂ as before

$$\frac{P^2_1 M_1}{a^3_1} = \frac{(8P_1)^2 2M_1}{(a_2)^3}$$ (4)

P₁ and M₁ cancel leaving 1/a₁³ = 128/a₂³ so a₂ = 5.039 a₁ = 5.039 times 1000000 km = 5000000 km.

5. (Harder) What is the value of the acceleration produced by a planet that is twice as massive and twice as large as Earth (hint F = m1a = G(m1m2/r²) remembering that the force is felt by you and relate the value of a to the value of g). What would you weight on this planet (assuming mass = 100kg).

So we use two of Newtons laws here

$$M_{\rm person} a_{\rm planet} = G \frac{M_{\rm person} M_{\rm planet}}{r_{\rm planet}^2}$$ (5)

The M $_{\rm person}$ cancel.

Now rearrange to have a constant value of one side

$$\frac{ a_{\rm planet} r^2_{\rm planet}}{M_{\rm planet}} = G$$ (6)

Now, know value for a on earth = g = 9.8 m s^-2. So

$$\frac{ a_{\rm planet} r_{\rm planet}^2}{M_{\rm planet}} = \frac{ a_{\rm earth} r_{\rm earth}^2}{M_{\rm earth}}$$ (7)

Told in Q that r $_{\rm planet}$ = 2r $_{\rm earth}$ M $_{\rm planet}$ = 2M $_{\rm earth}$ so

$$\frac{ a_{\rm planet} (2r_{\rm earth})^2}{2M_{\rm earth}} = \frac{ a_{\rm earth} r_{\rm earth}^2}{M_{\rm earth}}$$ (8)

rearth and Mearth cancel leaving

$$a_{\rm planet} 4/2 = a_{\rm earth}$$ (9)

so a for the planet is 1/2 a for earth = 1/2 9.8 = 4.9 m s^-2

What would you weight? F=ma = 100 4.9 = 490 N

6. (Sneaky one) Mercury orbits the sun with semi major axis 0.4 AU in 0.25 years. If the mass of Mercury was doubled, what would the period be if the distance was kept at 0.4 AU (Think of the assumptions we make).

Answer is the same. We make the assumption that the mass of mercury is tiny compared to the mass of the Sun so doubling the mass of mercury would NOT effect the calculation.

7. (Using small angle formula but same proportionality idea). At a distance of 12,000km, a planet appears to have an angular size of 4.4''. If the object is moved to 24,000km, what would the angular size be? Do this without calculating the linear size D by treating D*206265 as a constant (which it is the planet is unchanged.)

$$D = \frac{\alpha d}{206265}$$ (10)

so

$$\alpha d = 206265 D$$ (11)

Now D is fixed in the question (we don't change the physics size of the object) so as before $\alpha_1 d_1 = \alpha_2 d_2$ to if we double d (12,000km to 24,000 km) then we have to 1/2 the value of $\alpha_2$ to ans = 2.2''.

8. A planet is orbiting a star with semi-major axis a. Does it have a larger period if we double the semi-major axis or half the semi-major axis?

Doubling a) a₂ = 2a₁ and M₂ = M₁ so using

$$\frac{P^2_1 M_1}{a^3_1} = \frac{(P_2)^2 M_1}{(2a_1)^3}$$ (12)

we find P₂ = $\sqrt{8}$ P₁.

If we half a) then a₂ = 1/2 a₁ and then P₂ = 1/$\sqrt{8}$ P₁

so Period is larger if we double radius rather than half it (and by thinking about P² = a³ we could have got to this conclusion).