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Chapter 12: Problem 33

Suppose a new extrasolar planet is discovered. Its mass is double the mass of the Earth, but it has the same density and spherical shape as the Earth. How would the weight of an object at the new planet's surface differ from its weight on Earth?

Short Answer

Expert verified
Answer: To find the difference in weight, we need to calculate the ratio between the gravitational accelerations of the new planet and Earth (g_p/g_e). After substituting the values of M_p, R_p, M_e, and R_e in the formula, the ratio simplifies to: g_p/g_e = 4/3 This means that the weight of an object on the surface of the new planet would be 4/3 (or about 1.33 times) its weight on Earth.

Step by step solution

01

Recall the definition of density

Recall that the density of an object (\(\rho\)) is its mass (M) divided by its volume (V): \(\rho = \frac{M}{V}\). Since the density of our planet is the same as that of Earth, we can set up a proportionality between the planet's densities, masses, and volumes.
02

Set up an equation based on density

Divide the mass of the new planet (2M) by the mass of Earth (M) and the volume of the new planet (Vp) by the volume of Earth (Ve): \(\frac{2M}{M} = \frac{V_p}{V_e}\). From this, we can find the relationship between the volume of the new planet and Earth.
03

Solve for the new planet's volume

From Step 2, we have \(\frac{V_p}{V_e} = 2\). So, the volume of the new planet (\(V_p\)) is two times the volume of Earth (\(V_e\)): \(V_p = 2V_e\).
04

Find the relationship between the new planet's radius and Earth's radius

Recall that the volume of a sphere is given by \(V = \frac{4}{3}\pi R^3\). Since we know the volume of the new planet is double the volume of Earth, we can set up the following equation: \(2V_e = V_p\). Substituting the volume formulas for Earth and the new planet, we get: \(2\left(\frac{4}{3}\pi R_e^3\right) = \frac{4}{3}\pi R_p^3\), where \(R_p\) is the radius of the new planet and \(R_e\) is the radius of Earth. Divide both sides by \(\frac{4}{3}\pi\): \(2 R_e^3 = R_p^3\).
05

Solve for the new planet's radius

Now we solve for \(R_p\) in terms of \(R_e\): \(R_p = \sqrt[3]{2 R_e^3}\).
06

Calculate the gravitational acceleration

Now we have a relationship between \(M_p\) and \(R_p\): \(M_p = 2M_e\) and \(R_p = \sqrt[3]{2 R_e^3}\). We substitute these values into the formula for the gravitational acceleration (\(g_p = \frac{G\cdot M_p}{R_p^2}\)) for the new planet. For Earth (\(g_e = \frac{G\cdot M_e}{R_e^2}\)).
07

Compare weights

Since weight is given by W = mg, we can calculate the ratio of the weights on the new planet and Earth as: \(\frac{W_p}{W_e} = \frac{m \cdot g_p}{m \cdot g_e} = \frac{g_p}{g_e}\). Now substitute the gravitational acceleration expressions obtained in Step 6 and simplify the ratio. The resulting ratio will give us the relationship between the weights of an object on the new planet and Earth.

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Most popular questions from this chapter

You have been sent in a small spacecraft to rendezvous with a space station that is in a circular orbit of radius \(2.5000 \cdot 10^{4} \mathrm{~km}\) from the Earth's center. Due to a mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! You do not apply forward thrust in an attempt to chase the station; that would be fatal folly. Instead, you apply a brief braking force against the direction of your motion, to put you into an elliptical orbit, whose highest point is your present position, and whose period is half that of your present orbit. Thus, you will return to your present position when the space station has come halfway around the circle to meet you. Is the minimum radius from the Earth's center-the low point \(-\) of your new elliptical orbit greater than the radius of the Earth \((6370 \mathrm{~km})\), or have you botched your last physics problem?

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