Consider a planet whose mass and radius both are one half to that of Earth. An object of weight W on Earth will weigh ________ on that planet.
2 W
Calculating Weight on a Different Planet
The weight of an object is essentially the force of gravity acting on it. This force depends on the mass of the object and the gravitational acceleration at the object's location.
Understanding Weight and Gravity
Weight (\(W\)) is given by the formula:
\(W = m \cdot g\)
where:
- \(m\) is the mass of the object (which remains constant regardless of location).
- \(g\) is the acceleration due to gravity at that location.
The acceleration due to gravity (\(g\)) on the surface of a planet with mass \(M\) and radius \(R\) is given by:
\(g = \frac{GM}{R^2}\)
where \(G\) is the universal gravitational constant.
Applying the Formulas to Earth and the Planet
Let's denote the properties of Earth with subscript 'e' and the properties of the new planet with subscript 'p'. The mass of the object is \(m\).
On Earth, the weight of the object is given as \(W\). Using the formula:
\(W_e = m \cdot g_e = W\)
The acceleration due to gravity on Earth is:
\(g_e = \frac{GM_e}{R_e^2}\)
On the new planet, the weight of the object is:
\(W_p = m \cdot g_p\)
The acceleration due to gravity on the new planet is:
\(g_p = \frac{GM_p}{R_p^2}\)
Using the Given Information
We are told that the planet's mass and radius are both one half of Earth's. So:
- Mass of the planet, \(M_p = 0.5 M_e\)
- Radius of the planet, \(R_p = 0.5 R_e\)
Calculating Gravitational Acceleration on the Planet
Let's substitute the given values of \(M_p\) and \(R_p\) into the formula for \(g_p\):
\(g_p = \frac{G(0.5 M_e)}{(0.5 R_e)^2}\)
\(g_p = \frac{G(0.5 M_e)}{0.25 R_e^2}\)
\(g_p = \frac{0.5}{0.25} \cdot \frac{GM_e}{R_e^2}\)
\(g_p = 2 \cdot \frac{GM_e}{R_e^2}\)
We know that \(g_e = \frac{GM_e}{R_e^2}\), so we can write:
\(g_p = 2 g_e\)
This means the acceleration due to gravity on the new planet is twice that on Earth.
Calculating the Weight on the Planet
Now we can find the weight of the object on the planet:
\(W_p = m \cdot g_p\)
Substitute \(g_p = 2 g_e\):
\(W_p = m \cdot (2 g_e)\)
\(W_p = 2 \cdot (m \cdot g_e)\)
Since \(W_e = m \cdot g_e = W\), we have:
\(W_p = 2 W\)
So, an object weighing \(W\) on Earth will weigh \(2W\) on the planet.
Summary of Calculation
| Parameter | Earth | Planet | Relation |
|---|---|---|---|
| Mass | \(M_e\) | \(M_p\) | \(M_p = 0.5 M_e\) |
| Radius | \(R_e\) | \(R_p\) | \(R_p = 0.5 R_e\) |
| Gravity | \(g_e = \frac{GM_e}{R_e^2}\) | \(g_p = \frac{GM_p}{R_p^2}\) | \(g_p = 2 g_e\) |
| Weight | \(W_e = m g_e = W\) | \(W_p = m g_p\) | \(W_p = 2 W\) |
The weight of the object on that planet will be \(2W\).
Revision Table: Weight and Gravity Concepts
| Concept | Definition | Formula | Dependence |
|---|---|---|---|
| Weight (W) | Force of gravity on an object | \(W = m g\) | Mass of object (m), acceleration due to gravity (g) |
| Mass (m) | Amount of matter in an object | - | Constant for a given object |
| Acceleration due to gravity (g) | Acceleration experienced by an object due to gravitational force | \(g = \frac{GM}{R^2}\) | Gravitational constant (G), Mass of planet (M), Radius of planet (R) |
Additional Information: Factors Affecting Weight
Weight is not an intrinsic property of an object like mass is. It changes depending on the gravitational field it is in. The main factors influencing an object's weight are:
- Mass of the celestial body: A more massive planet exerts a stronger gravitational pull, leading to higher gravitational acceleration and thus greater weight for the same object.
- Radius of the celestial body: Gravitational acceleration decreases with the square of the distance from the center of the body. On the surface, this distance is the radius. So, a larger radius (for the same mass) means lower gravity and lower weight.
- Altitude: As an object moves away from the surface of a planet, the distance from the center increases, and the gravitational acceleration decreases, resulting in lower weight.
- Rotation of the planet: The rotation of a planet creates a centrifugal effect, which slightly reduces the apparent weight of objects at the surface, especially near the equator. However, this effect is usually small compared to the effect of mass and radius.
In this problem, both the planet's mass and radius are different from Earth's, and their combined effect results in a change in weight.
Similar Questions
"Pathfinder", an American Spacecraft, landed on the planet _______ in 1997.
- A/an ______ circle is a circular centre line on a drawing containing the centres of holes about a common centre.
- Which of the following is an example of a third-class lever?
_______ is the distance from the pitch circle to the bottom of the tooth space in a gear.
In a Class 3 lever, the effort and load move:
Important Questions from Miscellaneous
A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s2 and neglecting air resistance will give :
A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be:
Directions: Each item in this section consists of a sentence with an underlined word followed by four words (a), (b), (c), and (d). Select the option that is opposite in meaning to the underlined word and mark your response in your Answer Sheet accordingly.
The deluge affected the population.The major source of vitamins and minerals for vegetarians is
Which of the following statements about the Deccan Riots Commission is/are correct?
1. The Commission did not hold enquiries in the districts which were not affected.
2. The Commission did record the statements of ryots, sahukars and eye-witnesses.
Select the correct answer using the code given below:
