Question

A couple of astronauts agree to rendezvous in space after hours. Their plan is to let gravity bring them together. One of them has a mass of 65 kg and the other a mass of 72 kg, and they start from rest 20.0 m apart. (a) Make a free-body diagram of each astronaut, and use it to find his or her initial acceleration. As a rough approximation, we can model the astronauts as uniform spheres. (b) If the astronauts’ acceleration remained constant, how many days would they have to wait before reaching each other? (Careful! They both have acceleration toward each other.) (c) Would their acceleration, in fact, remain constant? If not, would it increase or decrease? Why?

Answer 1

Answer:

The answers to the questions are;

(a) The acceleration of the first astronaut is a₁ is 1.146 × 10⁻¹¹m·s⁻²

    The acceleration of the second astronaut is a₂ = 1.0347×10⁻¹¹m·s⁻²

(b) If the astronauts’ acceleration remained constant they would have to wait for 15.67 days before reaching each other.

(c) The acceleration would not remain constant but would increase as they get closer because a ∝ $\frac{1}{r^2}$.

Explanation:

To solve the question, we note

Firstly Newton's law of universal gravitation is given by

F₁ = F₂ = F = G×$\frac{m_1*m_2}{r^2}$

Where

m₁ = Mass of the first astronaut = 65 kg

m₂ = Mass of the second astronaut = 72 kg

F₁ = Force pulling the first astronaut to the second astronaut

F₂ = Force pulling the second astronaut to the first astronaut

G = Universal gravitational constant = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

r = Distance between the two astronauts = 20 m

Therefore F = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²× $\frac{65 kg*72 kg}{(20 m)^2}$ = 7.45 × 10⁻¹⁰m·kg·s⁻²

= 7.45 × 10⁻¹⁰ N

From Newton's second law, acceleration is directly proportional to the net force that produces it.

Therefore

F₁ = F₂ = F ∝ a

and F = m×a

For F₁ we have

F₁ = m₁×a₁ = 65 kg × a₁  = 7.45 × 10⁻¹⁰ N

a₁  = (7.45 × 10⁻¹⁰ N)/(65 kg) = 1.146 × 10⁻¹¹ N/kg = 1.146 × 10⁻¹¹ m·kg·s⁻²/kg

= 1.146 × 10⁻¹¹m·s⁻²

The acceleration of the first astronaut is a₁ = 1.146 × 10⁻¹¹m·s⁻²

The acceleration of the second astronaut is given by

For F₂ we have a₂ = (7.45 × 10⁻¹⁰ N)/(72 kg) = 1.0347×10⁻¹¹m·s⁻²

The acceleration of the second astronaut is a₂ = 1.0347×10⁻¹¹m·s⁻²

Therefore the average acceleration of the two astronauts is

$a_{average} = \frac{a_1+a_2}{2}$ =$\frac{1.0347*10^{-11}+1.146*10^{-11}}{2}$ =  1.09×10⁻¹¹m·s⁻².

(b)

Here we have from the laws of motion

v² = u² + 2×a×s

s = u·t +$\frac{1}{2}$·a·t²

Where:

v = Final velocity

u = Initial velocity = u₁, u₂ = 0

a = Acceleration = a₁, a₂

s = Distance traveled = s₁, s₂

t = time

To find the time, we have

Since the astronauts will both be travelling some distance s₁ and s₂ respectively at the same time t, we have

s = u·t +$\frac{1}{2}$·a·t² = 0·t +$\frac{1}{2}$·a·t² = $\frac{1}{2}$·a·t²

s₁ =  $\frac{1}{2}$·a₁·t² and s₂ =  $\frac{1}{2}$·a₂·t²

and s₁ + s₂ = 20.0 m

Therefore  $\frac{1}{2}$·a₁·t² +  $\frac{1}{2}$·a₂·t² = 20.0 m

⇒  $\frac{1}{2}$×1.146 × 10⁻¹¹m·s⁻²×t² +  $\frac{1}{2}$×1.0347×10⁻¹¹m·s⁻²×t² = 20.0 m

1.09×10⁻¹¹m·s⁻²×t² = 20.0 m

t² = (20.0 m)÷(1.09×10⁻¹¹m·s⁻²) = 1834125312301 s²

t =$\sqrt{1834125312301 s^2}$ = 1354298.8 s

But 1 day has 86400 seconds

Therefore time = (1354298.8 s)×1 day/(86400 s) = 15.67 days

Time to reach each other = 15.67 days.

(c) The acceleration would not remain constant but would increase as they get closer.

This is because the acceleration is given by

    F = m×a = G×$\frac{m_1*m_2}{r^2}$

That is the acceleration is inversely proportional to the square of the distance between the two astronauts

That is a ∝ $\frac{1}{r^2}$

therefore the acceleration will increase as the distance between the two astronauts decreases.