A body has a weight W on the surface of Earth. What is its weight on a planet whose mass is 15 times that of Earth and a radius that is 4 times that of the earth?
15/16 W
Understanding Weight on Different Celestial Bodies
The weight of an object is essentially the force of gravity exerted on it by a celestial body, such as Earth or a planet. This force depends on the mass of the celestial body, the mass of the object, and the distance between the center of the celestial body and the object.
Formula for Gravitational Force and Weight
According to Newton's Law of Universal Gravitation, the force of gravity (which is the weight, W) between two objects is given by the formula:
$$W = G \frac{M m}{R^2}$$
Where:
- \(G\) is the universal gravitational constant.
- \(M\) is the mass of the larger body (Earth or the planet).
- \(m\) is the mass of the smaller body (the object).
- \(R\) is the distance between the centers of the two bodies. On the surface, this distance is the radius of the celestial body.
Weight on Earth
Let \(W_E\) be the weight of the body on the surface of Earth. Let \(M_E\) be the mass of Earth and \(R_E\) be the radius of Earth. The problem states that the weight on Earth is \(W\).
So, the weight on Earth can be written as:
$$W_E = G \frac{M_E m}{R_E^2} = W$$
Applying the Formula to the Planet
Now, let's consider the weight of the same body on the distant planet. Let \(W_P\) be the weight on the planet, \(M_P\) be the mass of the planet, and \(R_P\) be the radius of the planet.
We are given the following information about the planet compared to Earth:
- The mass of the planet is 15 times that of Earth: \(M_P = 15 M_E\).
- The radius of the planet is 4 times that of Earth: \(R_P = 4 R_E\).
The weight of the body on the planet is given by:
$$W_P = G \frac{M_P m}{R_P^2}$$
Calculating the Weight on the Planet
Substitute the given values for \(M_P\) and \(R_P\) in terms of \(M_E\) and \(R_E\) into the formula for \(W_P\):
$$W_P = G \frac{(15 M_E) m}{(4 R_E)^2}$$
Simplify the denominator:
$$W_P = G \frac{15 M_E m}{16 R_E^2}$$
Now, we can rearrange the expression to relate it back to the weight on Earth, \(W_E = W\):
$$W_P = \frac{15}{16} \left( G \frac{M_E m}{R_E^2} \right)$$
Since \(G \frac{M_E m}{R_E^2} = W_E = W\), we can substitute \(W\) into the equation:
$$W_P = \frac{15}{16} W$$
Therefore, the weight of the body on the planet is \(\frac{15}{16}\) times its weight on Earth.
Summary of Calculation
| Quantity | On Earth | On Planet |
|---|---|---|
| Mass of body | \(m\) | \(m\) |
| Mass of celestial body | \(M_E\) | \(M_P = 15 M_E\) |
| Radius of celestial body | \(R_E\) | \(R_P = 4 R_E\) |
| Weight | \(W_E = G \frac{M_E m}{R_E^2} = W\) | \(W_P = G \frac{M_P m}{R_P^2} = G \frac{15 M_E m}{(4 R_E)^2} = G \frac{15 M_E m}{16 R_E^2} = \frac{15}{16} \left(G \frac{M_E m}{R_E^2}\right) = \frac{15}{16} W\) |
Conclusion on Planet Weight
The weight of the body on the planet is \(\frac{15}{16} W\).
Revision Table: Gravity Concepts
| Concept | Description | Formula |
|---|---|---|
| Weight (W) | Force of gravity on an object near a celestial body's surface. | \(W = mg\) |
| Acceleration due to gravity (g) | Acceleration experienced by an object due to gravity. | \(g = G \frac{M}{R^2}\) |
| Universal Gravitational Constant (G) | A fundamental constant in the law of universal gravitation. | \(G \approx 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2\) |
Additional Information: Factors Affecting Weight
The weight of an object is not a fixed property of the object itself, unlike its mass. Weight depends on the gravitational field it is in. The strength of the gravitational field, and therefore the acceleration due to gravity (\(g\)), depends on:
- Mass of the celestial body (M): A more massive body creates a stronger gravitational pull, leading to greater weight. Weight is directly proportional to the mass of the celestial body.
- Distance from the center (R): The gravitational force decreases rapidly with distance. Weight is inversely proportional to the square of the distance from the center of the celestial body. For an object on the surface, this distance is the radius.
So, even though the object's mass (\(m\)) remains constant, its weight changes when it moves from Earth to a planet with different mass and radius.
Similar Questions
The acceleration due to gravity on the Moon is (1/6) of that on the Earth. Hence, an object weighing 12 N on the Earth will weigh ________ on the Moon.
Fill in the blank with the most appropriate option.
The Universal Constant of Gravitation is ________.
Which of the following statements is/are INCORRECT?
A. The ratio of the force of gravitation between two masses, m1 and m2, kept at a distance R on the earth and on the moon is 1:1.
B. Nm2/kg2 is the SI unit of G.
C. The value of G depends on the distance between the bodies.
D. The value of G depends on the masses of the bodies.Calculate the work done by the force of gravity when a satellite moves in an orbit of radius 40,000 km around the earth.
What is the value of acceleration due to gravity on the surface of the earth?
The value of g on the moon is 1/6 th of the value of g on the earth. If a man can jump 1.5 m high on the earth, on the moon, he can jump up to a height of:
Consider a hypothetical planet whose mass and radius are both half that of Earth. If g is the acceleration due to gravity on the surface of Earth, the acceleration due to gravity on the planet will be:
A 5 kg object is raised through a height of 4 m. The Work done by the force of gravity acting on the object is (take g = 10 m/s 2):
Consider a planet whose mass and radius are both twice the mass and radius of Earth. The acceleration due to gravity on the surface of the planet is n times that on Earth. The value of n is:
Important Questions from Gravity
Which of the following law states that, "The force between two objects is directly proportional to the product of their masses?"
Which of the following statements about the movement of planets is true?
A. A planet's orbit is elliptical with the Sun at one of two focal points.
B. The orbit of a planet is circular with the sun in the center.
C. The orbit of a planet is elliptical with another planet in one of the two center-points.
D. The orbit of a planet is circular with another planet in the center.
If the mass of a person is 60 kg on the surface of earth then the same person’s mass on the surface of the moon will be:
The centripetal force required to keep the moon in its orbit is provided by which force?
How is the acceleration due to gravity denoted?
