2023 MT Physics Structured Questions

Question 1

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As shown in the figure, a satellite S of mass $m$ is orbiting Earth of mass $M_{E}$ in a circular path of radius $r$ . It is parked at a geostationary orbit, which means that it has the same period as Earth’s rotation (i.e. 24 h) and is able to maintain a constant position directly above a fixed point on Earth near the equator. The satellite has a mass of 700 kg. The mass of the Earth is $5.972 \times 10^{24}$ kg.

a. Determine
i. the radius of this orbit, and
ii. the general formula for the total energy of a satellite in terms of its mass $m$ and orbital radius $r$ round Earth. You can include any other physical constants needed in your formula. Calculate the numerical value of the total energy for this satellite.

b. After a few years, the satellites may no longer function nor orbit correctly. One method of disposing dysfunctional geo-stationary satellites is to cause the satellite to be ‘de-orbited’ and made to crash into uninhabited parts of the Earth. Using this method, a satellite of mass 700 kg began to make its descent at 90 km above the Earth’s surface for 15 min just before impact. It stays at a constant speed all the way from initial descent to just before crashing. Assuming that all energy losses are converted only to heat, estimate the average rate of heat generated on the descent. (Hint: 90 km is considered to be close to Earth’s surface and the gravitational field strength can be treated as constant.)

c. Another method of disposing dysfunctional low flying satellite is by using the remaining fuel to propel them into a ‘graveyard orbit’—another orbit which is hundreds of kilometres from the geo-stationary orbit. In this method, the satellite is first given a velocity boost of about 11 m s $^{- 1}$ in its original direction of motion and another boost later to keep the final orbit circular. Is the radius of this graveyard orbit larger or smaller compared to the original orbit? With reference to the total energy of the satellite, explain your answer.

Solutions

Part a: Geostationary Orbit Radius and Total Energy

For a satellite to maintain a stable circular orbit, the gravitational force exerted by the Earth must provide the necessary centripetal force required for circular motion.

i. Determining the Radius of the Orbit ( $r$ )

We equate the gravitational force ( $F_{g}$ ) to the centripetal force ( $F_{c}$ ):

F_{g} = F_{c}

\frac{G M_{E} m}{r^{2}} = m a_{c}

Since the satellite is in a circular orbit, the centripetal acceleration ( $a_{c}$ ) can be expressed in terms of its angular speed ( $ω$ ) or its orbital period ( $T$ ): $a_{c} = r ω^{2}$ . Since we are given the period $T$ , let's use the relationship $ω = \frac{2 π}{T}$ .

\frac{G M_{E} m}{r^{2}} = m r {(\frac{2 π}{T})}^{2}

Notice that the satellite's mass ( $m$ ) cancels out—a great reminder that the orbital radius only depends on the mass of the central body ( $M_{E}$ ) and the orbital period ( $T$ ).

Now, let's solve for the radius ( $r$ ):

G M_{E} = r^{3} {(\frac{2 π}{T})}^{2}

r^{3} = \frac{G M_{E} T^{2}}{4 π^{2}}

r = \sqrt[3]{\frac{G M_{E} T^{2}}{4 π^{2}}}

Given Values:

Gravitational Constant, $G \approx 6.674 \times 10^{- 11} N \cdot m^{2} / {kg}^{2}$
Mass of Earth, $M_{E} = 5.972 \times 10^{24} kg$
Period of orbit (Geostationary), $T = 24 h$ . We must convert this to seconds:

* * C a l c u l a t i o n : * *

r^3 = \frac{(6.674 \times 10^{-11} \text{ N}\cdot\text{m}^2/\text{kg}^2) \times (5.972 \times 10^{24} \text{ kg}) \times (86,400 \text{ s})^2}{4\pi^2}

r^3 \approx 7.537 \times 10^{22} \text{ m}^3

r \approx 4.223 \times 10^7 \text

The radius of the geostationary orbit is approximately **$4.22 \times 10^7 \text{ m}$** (or $42,230 \text{ km}$). --- #### ii. General Formula and Numerical Value for Total Energy The **total mechanical energy** ($E_{total}$) of a satellite in orbit is the sum of its **kinetic energy** ($K$) and its **gravitational potential energy** ($U$).

E_{total} = K + U

T h e g r a v i t a t i o n a l p o t e n t i a l e n e r g y i s g i v e n b y :

U = - \frac{G M_E m}

T h e k i n e t i c e n e r g y i s :

K = \frac{1}{2} m v^2

F r o m P a r t (i), w e k n o w $ \frac{G M_{E} m}{r^{2}} = \frac{m v^{2}}{r} $, s o $ m v^{2} = \frac{G M_{E} m}{r} $ . T h e r e f o r e, t h e k i n e t i c e n e r g y i s :

K = \frac{1}{2} m v^2 = \frac{1}{2} \left(\frac{G M_E m}{r}\right) = \frac{G M_E m}

T h e G e n e r a l F o r m u l a f o r T o t a l E n e r g y i s :

E_{total} = K + U = \frac{G M_E m}{2r} + \left(-\frac{G M_E m}{r}\right)

E_{total} = - \frac{G M_E m}

T h e n e g a t i v e s i g n i n d i c a t e s t h a t t h e s a t e l l i t e i s * * b o u n d * * t o t h e E a r t h; y o u^{'} d n e e d t o a d d e n e r g y t o i t t o g e t i t t o e s c a p e . T h i s i s t h e * * v i r i a l t h e o r e m * * f o r a $ 1 / r $ p o t e n t i a l! * * C a l c u l a t i o n o f t h e N u m e r i c a l V a l u e : * * N o w, l e t^{'} s p l u g i n t h e n u m e r i c a l v a l u e s : - $ G \approx 6.674 \times 10^{- 11} N \cdot m^{2} / {kg}^{2} $ - $ M_{E} = 5.972 \times 10^{24} kg $ - $ m = 700 kg $ - $ r \approx 4.223 \times 10^{7} m $ (f r o m p a r t i)

E_{total} = - \frac{(6.674 \times 10^{-11}) \times (5.972 \times 10^{24}) \times (700)}{2 \times (4.223 \times 10^7)}

E_{total} \approx - \frac{2.790 \times 10^{17} \text{ J}\cdot\text{m}}{8.446 \times 10^7 \text{ m}}

E_{total} \approx -3.30 \times 10^9 \text

The total energy of the geostationary satellite is approximately **$-3.30 \times 10^9 \text{ J}$**. --- ## Part b: Average Rate of Heat Generated During Descent This part involves the **conservation of energy** with a non-conservative force (air resistance/drag) causing energy loss, which is converted to heat. The satellite maintains a constant speed, meaning $\Delta K = 0$. The **Work-Energy Theorem** states that the work done by all forces equals the change in kinetic energy:

W_{net} = \Delta K

T h e n e t w o r k i s t h e w o r k d o n e b y g r a v i t y ($ W_{g} $) p l u s t h e w o r k d o n e b y t h e n o n - c o n s e r v a t i v e f o r c e s ($ W_{n c} $, w h i c h c o n v e r t s t o h e a t i n t h i s c a s e) :

W_g + W_{nc} = \Delta K

S i n c e t h e s p e e d i s c o n s t a n t, $ Δ K = 0 $ .

W_{nc} = - W_g

A l s o, t h e w o r k d o n e b y g r a v i t y i s e q u a l t o t h e n e g a t i v e c h a n g e i n p o t e n t i a l e n e r g y : $ W_{g} = - Δ U $ .

W_{nc} = - (-\Delta U) = \Delta U

T h e e n e r g y l o s t a s h e a t i s e q u a l t o t h e m a g n i t u d e o f t h e w o r k d o n e b y t h e n o n - c o n s e r v a t i v e f o r c e s, $ Q = | W_{n c} | $ . S i n c e $ Δ U $ i s p o s i t i v e (p o t e n t i a l e n e r g y_{i} n c r e a s e s_{a} s y o u m o v e_{a} w a y_{f} r o m E a r t h, s o i t_{d} e c r e a s e s_{d} u r i n g d e s c e n t), t h e h e a t g e n e r a t e d $ Q $ i s e q u a l t o t h e d e c r e a s e i n g r a v i t a t i o n a l p o t e n t i a l e n e r g y, $ | Δ U | $ . * * E s t i m a t e t h e D e c r e a s e i n P o t e n t i a l E n e r g y ($ | Δ U | $) : * * T h e p r o b l e m s t a t e s t h a t $ 90 km $ i s c l o s e t o t h e E a r t h^{'} s s u r f a c e, s o w e c a n u s e t h e c o n s t a n t g r a v i t a t i o n a l f i e l d a p p r o x i m a t i o n, $ Δ U \approx m g Δ h $ . * * G i v e n V a l u e s : * * - M a s s o f s a t e l l i t e, $ m = 700 kg $ - G r a v i t a t i o n a l a c c e l e r a t i o n (a t E a r t h^{'} s s u r f a c e), $ g \approx 9.8 m / s^{2} $ - D e s c e n t d i s t a n c e, $ Δ h = 90 km = 90, 000 m $ - T i m e o f d e s c e n t, $ Δ t = 15 min = 15 \times 60 s = 900 s $ T h e h e a t g e n e r a t e d ($ Q $) i s t h e d e c r e a s e i n p o t e n t i a l e n e r g y :

Q = |\Delta U| = m g \Delta h

Q = (700 \text{ kg}) \times (9.8 \text{ m}/\text{s}^2) \times (90,000 \text{ m})

Q = 617,400,000 \text{ J} \approx 6.17 \times 10^8 \text

* * A v e r a g e R a t e o f H e a t G e n e r a t e d ($ P_{h e a t} $) : * * T h e a v e r a g e r a t e o f h e a t g e n e r a t e d i s t h e t o t a l h e a t g e n e r a t e d d i v i d e d b y t h e t i m e t a k e n :

P_{heat} = \frac{Q}{\Delta t}

P_{heat} = \frac{6.174 \times 10^8 \text{ J}}{900 \text{ s}}

P_{heat} \approx 686,000 \text

The average rate of heat generated during the descent is approximately **$6.86 \times 10^5 \text{ W}$** (or $686 \text{ kW}$). --- ## Part c: Graveyard Orbit Radius and Total Energy The satellite is given a **velocity boost** of $11 \text{ m/s}$ **in its original direction of motion**. This means its new kinetic energy ($K'$) will be greater than its original kinetic energy ($K$).

K' = \frac{1}{2} m (v + \Delta v)^2 > \frac{1}{2} m v^2 = K

S i n c e $ E_{t o t a l} = K + U $ a n d $ U $ (p o t e n t i a l e n e r g y) i s u n c h a n g e d i n i t i a l l y (a s t h e b o o s t i s a p p l i e d i n s t a n t a n e o u s l y a t t h e o r i g i n a l r a d i u s $ r $), t h e * * t o t a l e n e r g y ($ E_{t o t a l} $) m u s t i n c r e a s e * * :

E'{total} = K' + U > K + U = E

### Total Energy and Orbital Radius Recall the formula for the total energy from Part a(ii):

E_{total} = - \frac{G M_E m}

- S i n c e $ G $, $ M_{E} $, a n d $ m $ a r e p o s i t i v e, $ E_{t o t a l} $ i s a l w a y s n e g a t i v e f o r a b o u n d o r b i t . - T h e t o t a l e n e r g y $ E_{t o t a l} $ i s * * i n v e r s e l y p r o p o r t i o n a l * * t o t h e o r b i t a l r a d i u s $ r $ . B e c a u s e t h e i n i t i a l b o o s t * * i n c r e a s e s * * t h e t o t a l e n e r g y ($ E_{t o t a l}^{'} > E_{t o t a l} $), t h e n e w t o t a l e n e r g y i s a * * s m a l l e r n e g a t i v e n u m b e r * * (i . e ., l e s s n e g a t i v e, c l o s e r t o z e r o) . F o r $ E_{t o t a l}^{'} $ t o b e a s m a l l e r n e g a t i v e n u m b e r, t h e t e r m $ \frac{1}{r} $ m u s t b e s m a l l e r (o r $ r $ m u s t b e l a r g e r) .

\text{If } E'{total} > E \text{ (less negative)}, \text{ then } \left|E'{total}\right| < \left|E\right|

\text{Since } \left|E_{total}\right| \propto \frac{1}{r}, \text{ then } \frac{1}{r'} < \frac{1}{r} \implies r' > r

T h e r e f o r e, t h e r a d i u s o f t h e g r a v e y a r d o r b i t ($ r^{'} $) m u s t b e * * l a r g e r * * c o m p a r e d t o t h e o r i g i n a l o r b i t r a d i u s ($ r $) . * * E x p l a n a t i o n S u m m a r y : * * 1. A p p l y i n g a v e l o c i t y b o o s t i n t h e d i r e c t i o n o f m o t i o n * * i n c r e a s e s t h e s a t e l l i t e^{'} s k i n e t i c e n e r g y * * a n d, c o n s e q u e n t l y, i t s * * t o t a l m e c h a n i c a l e n e r g y ($ E_{t o t a l} $) * * . 2. T h e t o t a l e n e r g y o f a s a t e l l i t e i n a c i r c u l a r o r b i t i s g i v e n b y $ E_{t o t a l} = - \frac{G M_{E} m}{2 r} $ . 3. A n * * i n c r e a s e i n $ E_{t o t a l} $ * * m e a n s t h e t o t a l e n e r g y b e c o m e s a * * s m a l l e r (l e s s n e g a t i v e) n u m b e r * * . 4. F o r t h e v a l u e o f $ - \frac{G M_{E} m}{2 r} $ t o b e c o m e a s m a l l e r n e g a t i v e n u m b e r (c l o s e r t o z e r o), t h e d e n o m i n a t o r $ r $ * * m u s t i n c r e a s e * * . 5. T h u s, t h e g r a v e y a r d o r b i t i s * * l a r g e r * * t h a n t h e o r i g i n a l g e o s t a t i o n a r y o r b i t .

Question 1

Solutions

Part a: Geostationary Orbit Radius and Total Energy

i. Determining the Radius of the Orbit (r)

i. Determining the Radius of the Orbit ( $r$ )