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Planet A has a mass that is twice as large as the mass of planet B and a radius that is twice as large as the radius of planet B. Calculate the ratio of the gravitational field strength on planet A to that on planet B.

Short Answer

Expert verified
The ratio is 1/2.

Step by step solution

01

Write the Formula for Gravitational Field Strength

The gravitational field strength (\textbf{g}) at the surface of a planet is given by the formula \[ g = \frac{GM}{R^2} \] where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.
02

Define the Mass and Radius of Planet B

Let the mass of planet B be \( M_B \) and its radius be \( R_B \). Then, the gravitational field strength on planet B is:\[g_B = \frac{G M_B}{R_B^2}\].
03

Define the Mass and Radius of Planet A

Planet A has twice the mass of planet B and twice the radius of planet B. So, we can write:\[M_A = 2M_B\] and \[R_A = 2R_B\].
04

Calculate the Gravitational Field Strength on Planet A

Substitute the values for \( M_A \) and \( R_A \) into the formula for gravitational field strength:\[g_A = \frac{G M_A}{R_A^2} = \frac{G (2M_B)}{(2R_B)^2} = \frac{2G M_B}{4R_B^2} = \frac{G M_B}{2R_B^2}\].
05

Calculate the Ratio of Gravitational Field Strengths

The ratio of the gravitational field strength on planet A to that on planet B is:\[\frac{g_A}{g_B} = \frac{\frac{G M_B}{2R_B^2}}{\frac{G M_B}{R_B^2}} = \frac{1}{2}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Planetary Mass and Radius
Understanding how the mass and radius of a planet affect its gravitational field strength is crucial.

The mass of a planet (M) and its radius (R) directly influence the gravitational field strength.
Gravitational field strength (\textbf{g}) is essentially the force per unit mass experienced by an object in the planet's gravitational field.

According to the equation \( g = \frac{GM}{R^2} \), the field strength is proportional to the planet's mass.
However, it is inversely proportional to the square of the planet's radius.

This means that:
  • If the planet's mass increases, the gravitational field strength increases.
  • If the planet's radius increases, the gravitational field strength decreases.
In the exercise, planet A has twice the mass and twice the radius of planet B.
By substituting these values into the formula, one can understand how variations in mass and radius influence gravitational field strength.
Gravitational Constant
The gravitational constant (G) is a fundamental physical constant used in the calculation of gravitational forces.

It's value is approximately 6.674×10^{-11} N m²/kg².
Despite the planets' differing masses and radii, this value remains constant.

Including G in the formula \( g = \frac{GM}{R^2} \), ensures that gravitational calculations are accurate and consistent across different celestial bodies.
It's crucial to remember that while mass and radius can vary, G provides the steady foundation necessary for these computations.

For instance, if we didn't have a constant such as G, calculating the gravitational field strength for various planets would lack consistency.
Therefore, understanding the role of the gravitational constant helps in appreciating its importance in maintaining uniformity in our gravitational measurements and calculations.
Ratio Calculations
Calculating ratios helps to compare different quantities against each other. In this exercise, we determine the ratio of gravitational field strengths.

Given planets A and B, the gravitational field strength on planet B is given by \( g_B = \frac{G M_B}{R_B^2} \).
For planet A, with twice the mass and radius of planet B, its field strength is \( g_A = \frac{G (2M_B)}{(2R_B)^2} = \frac{2G M_B}{4R_B^2} = \frac{G M_B}{2R_B^2} \).

To find the ratio of the gravitational field strength on planet A to that on planet B, we use the formula:
\[ \frac{g_A}{g_B} = \frac{\frac{G M_B}{2R_B^2}}{\frac{G M_B}{R_B^2}} = \frac{1}{2} \]. This simplifies the comparison by canceling out common factors.

Thus, the gravitational field strength on planet A is half that of planet B.
Ratio calculations can greatly simplify complex problems by providing relative measures.
They offer insight into how varying a few parameters can drastically alter outcomes.

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Most popular questions from this chapter

A man of mass \(m\) stands in an elevator. Find the reaction force from the elevator floor on the man when: (a) the elevator is standing still; (b) the elevator moves up at constant speed \(v\) (c) the elevator accelerates down with acceleration a; (d) the elevator accelerates down with acceleration \(a=g\) (e) What happens when \(a>g ?\)

A ball rolls off a table with a horizontal speed \(2.0 \mathrm{m} \mathrm{s}^{-1} .\) If the table is \(1.3 \mathrm{m}\) high, how far from the table will the ball Jand?

Show by applying Newton's law of gravitation and the second law of mechanics that a satellite (or planet) in a circular orbit of radius \(R\) around the earth (or the sun) has a period (i.e. time to complete one revolution) given by $$T^{2}=\frac{4 \pi^{2} R^{3}}{G M}$$ where \(M\) is the mass of the attracting body (earth or sun). This is Kepler's third law.

Show that the escape speed from the surface of a planet of radius \(R\) can be written as \(v_{\mathrm{esc}}=\sqrt{2 g R},\) where \(g\) is the gravitational field strength on the planet's surface.

Make velocity-time sketches (no numbers are necessary on the axes) for the following motions. (a) A ball is dropped from a certain height and bounces off a hard floor. The speed just before each impact-with the floor is the same as the speed just after impact. Assume that the time of contact with the floor is negligibly small. (b) A cart slides with negligible friction along a horizontal air track. When the cart hits the ends of the air track it reverses direction with the same speed it had right before impact. Assume the time of contact of the cart and the ends of the air track is negligibly small. (c) A person jumps from a hovering helicopter. After a few seconds she opens a parachute. Eventually she will reach a terminal speed and will then land.

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