A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other objects i

Q: Why does the weight become exactly zero in the middle of the journey?

Weight is the net gravitational force acting on the body. As you move away from Earth and towards Mars, the Earth pulls you back while Mars pulls you forward. At a specific location called the 'neutral point', these two opposing forces are equal in magnitude. They cancel each other out, making the net force (and thus your weight) zero.

Q: Why isn't the graph a simple straight line connecting 1000 N and 400 N?

Gravitational force follows an inverse square law ($F \propto 1/r^2$). This means the force doesn't decrease linearly with distance; it drops off much faster initially and then levels out. This non-linear relationship creates the curved shape of the graph.

Q: Does the mass of the passenger change during the journey?

No, mass is an intrinsic property of an object that represents the amount of matter it contains. It remains a constant $100 \text{ kg}$ regardless of location. Only weight, which depends on local gravity, changes.

Q: What if we consider the Sun's gravity? Wouldn't that affect the weight?

In reality, yes, the Sun's gravity plays a massive role in interplanetary travel. However, the problem explicitly instructs us to 'neglect all other objects in the sky' to isolate and test your understanding of the two-body gravitational interaction between Earth and Mars.

Q: Why does Curve IV go below the zero line?

Curve IV implies a 'negative weight'. While gravitational force is a vector and can have a negative sign depending on your coordinate system, weight is typically measured as a magnitude (like what a spring balance would read), which cannot be negative. Therefore, Curve IV is physically incorrect for this context.

Question

A person whose mass is $100 	ext{ kg}$ travels from Earth to Mars in a spaceship. Neglect all other objects in sky and take acceleration due to gravity on the surface of the Earth and Mars as $10 	ext{ m/s}^2$ and $4 	ext{ m/s}^2$, respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.

Options:

(A) III

(B) I

(C) IV

(D) II

Sigma Percentile · Accepted Answer

### The Setup: Leaving Earth

Imagine you are an astronaut embarking on an epic interplanetary journey from Earth to Mars. Your mass is a constant $100 	ext{ kg}$. Mass is an intrinsic property of matter; it doesn't change whether you are on Earth, in deep space, or on Mars. However, your **weight** is a different story. Weight is the gravitational force exerted on you, given by the equation:

$$W = mg$$

Before you take off, you are standing on the surface of the Earth. The acceleration due to gravity here is $g_E = 10 	ext{ m/s}^2$. Plugging this into our equation, your initial weight is:

$$W_E = 100 	imes 10 = 1000 	ext{ N}$$

### The Interplanetary Tug of War

As your spaceship blasts off and travels further away from Earth, the Earth's gravitational pull on you begins to weaken. This is because gravity follows an **inverse square law** ($g \propto \frac{1}{r^2}$).

Simultaneously, as you get closer to Mars, its gravitational pull on you starts to increase. Throughout the journey, you are caught in a cosmic tug of war. The Earth is pulling you back, and Mars is pulling you forward. The net gravitational field you experience is the vector sum of these two opposing fields.

### The Neutral Point: True Weightlessness

Because you are moving from a stronger gravitational field (Earth) to a weaker one (Mars), there must be a specific location in space where the backward pull of the Earth perfectly balances the forward pull of Mars.

At this exact location, known as the **neutral point**, the net gravitational acceleration becomes zero ($g_{net} = 0$). Consequently, your weight drops to absolute zero:

$$W_{net} = 100 	imes 0 = 0 	ext{ N}$$

This is the moment of true weightlessness during your journey.

### Arrival: A Lighter Step on Mars

After passing the neutral point, the gravitational pull of Mars becomes dominant. As you approach the Martian surface, your weight begins to increase again. Upon landing, the acceleration due to gravity on Mars is $g_M = 4 	ext{ m/s}^2$. Your final weight is:

$$W_M = 100 	imes 4 = 400 	ext{ N}$$

Looking at the given graphs, we need a curve that starts at $1000 	ext{ N}$, dips all the way down to $0 	ext{ N}$ at the neutral point, and then rises to end at $400 	ext{ N}$. **Curve III** is the only graph that perfectly captures this physical reality. The curve is non-linear (not a straight line) precisely because of the inverse square nature of Newton's Law of Universal Gravitation.

Select Answer:

Visualized Solution

\text{The Interplanetary Journey}

\text{Defining Weight}

\text{Initial Weight on Earth}

\text{Final Weight on Mars}

\text{The Tug of War}

\text{The Neutral Point}

\text{Identifying the Curve}

\text{Non-Linear Variation}

The Sigma Insight: Acceleration due to Gravity and its Variation

The Setup

The Interplanetary Tug of War

The Neutral Point

Arrival

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