**Imagine if the Moon doubles in mass, which of the following is likely to happen?**

**Each month will be longer than 30 days.****Nights will be a lot brighter since the Moon is nearer to the Earth****Each day will be longer since the period of the Moon increases.****The Moon continues to revolve around the Earth in the same orbit.**

**Show/Hide Answer**

T^{2} = kr^{3} therefore period and orbital radius is independent of mass.

Answer: 4

**Which quantity is not necessarily the same for satellites that are in geostationary orbits around the Earth?**

**Angular velocity****Kinetic energy****Centripetal acceleration****Orbital period**

**Show/Hide Answer**

Period of geostationary satellite’s orbit is the same as Earth (24 hours) and hence, its angular velocity is also constant. It can also be found that the radius of the geostationary orbit is fixed using R^{3}∝T^{2}. Hence centripetal acceleration is also constant. The only quantity that varies is the kinetic energy because it depends on the mass of the individual satellites.

Answer: 2

**Mars is known to possess two satellites, Phobos and Deimos. The former is at a distance of 9,500 km from the centre of Mars and the latter at a distance of 24,500 km. Find the ratio of the period of Phobos to that of Deimos in their revolutions around Mars.**

**Show/Hide Answer**

$\frac{T_{phobos}}{T_{deimos}} = \sqrt{\frac{r^{3}_{phobos}}{r^{3}_{deimos}}} = \sqrt{\frac{9500^{3}}{24500^{3}}} = 0.24$

Answer: 0.24

**Assuming the Earth to be a uniform sphere rotating about an axis through the poles, the apparent weight of a body at the Equator compared with it’s apparent weight at the poles would be**

**greater, because the angular velocity of the Earth is greater at the Equator than at the pole****greater, because the weight at the Equator is given by the sum of the gravitational attraction of the Earth and the centripetal force due to the circular motion of the body****smaller, because the gravitational attraction of the Earth must provide both the weight and the centripetal force due to the circular motion of the body****smaller, because the gravitational attraction at the pole is greater than that at the Equator**

**Show/Hide Answer**

The Earth spins about the axis through which its poles pass. A person at the Equator undergoes circular motion while a person at the poles does not.

Answer: 3

**a) The Moon is orbiting at a certain height above the Earth. The Earth exerts a force of 1.99 x 10 ^{20} N on the Moon.**

**Given: mass of the Earth = 5.98 x 10 ^{24} kg**

**mass of the Moon = 7.35 x 10**

^{22}kg**radius of the Earth = 6.38 x 10**

^{6}m**i) Find the height above the Earth’s surface where the Moon is orbiting.**

**ii) Using Newton’s Second Law of motion, calculate the linear velocity of the Moon.**

**iii) Calculate the centripetal acceleration of the Moon.**

**b) Most artificial satellites are placed in Low Earth Orbits (LEO), at a height typically around 200 km – 1200 km above the Earth’s surface. A 700 kg satellite is in the Low Earth Orbit at a height of 1200 km above the Earth’s surface.**

**i) What is the change in gravitational potential energy of the satellite before it is launched and when it is in orbit?**

**ii) Calculate the period of the satellite. Express your answer in hours.**

**iii) Satellites in LEO are used in telecommunications. State one difference between the LEO and geostationary satellites, and an advantage of using LEO satellites in telecommunications.**

**Show/Hide Answer**

(a) i) F = G(M

_{E}M

_{M}/r

^{2})

r = 3.8382 x 10

^{8}m

Height above Earth’s surface = 3.8382 x 10^{8} – 6.38 x 10^{6}

= 3.77 x 10^{8} m

ii) By Newton’s second law of motion,

F_{g} = F_{c}

v = (GM_{E}/r)^{1/2}

v = 1.02 x 10^{3} m/s

iii) a = v^{2}/r

a = 2.71 x 10^{-3} m s^{-2}

b) i) Change in GPE = U_{o} – U_{s}, where o is orbit, s is surface

= m(ϕ_{o} – ϕ_{s})

= 6.93 x 10^{9} J

ii) By Newton’s second law of motion,

F_{g} = mrω^{2}

F_{g} = mr(2π/T)^{2}

T = 6566s

T = 1.82 hrs.

iii)LEO satellites have a shorter orbital period as compared to geostationary satellites which have a period of 24 hrs.

LEO satellites are nearer to the Earth and so the resolution of the images received will be higher.

actually, i just realized i forgot to change them to SI units… Thanks anyway!

Nice. 🙂

for b) i) , i couldn’t manage to get your answer. Can you clarify? thanks a bunch!