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The planet zeta has 2 times the gravitational field strength and sane mass as the earth , how does the radius of zeta compare with the radius of the earth

User Bu Saeed
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2 Answers

4 votes

Final answer:

The radius of planet Zeta is approximately √2 times the radius of Earth.

Step-by-step explanation:

The gravitational field strength depends on the mass and radius of a planet. In this case, planet Zeta has 2 times the gravitational field strength and the same mass as the Earth. This means that the radius of Zeta must be smaller than the radius of Earth. Let's assume the radius of Earth is represented by RE and the radius of Zeta is represented by RZ.

According to the given information, the mass of Zeta (MZ) is equal to the mass of Earth (ME). The gravitational field strength of Zeta (gZ) is 2 times the gravitational field strength of Earth (gE). Using the formula for gravitational field strength, g = GM / R², we can set up the equation for Zeta as: gZ = (G * MZ) / RZ² and for Earth as: gE = (G * ME) / RE².

Since the mass of Zeta (MZ) is equal to the mass of Earth (ME), we can set up the equation as: gZ = (G * ME) / RZ², where gZ = 2 * gE. Rearranging the equation and substituting values, we get: (G * ME) / RZ² = 2 * [(G * ME) / RE²].

Cross-multiplying and simplifying, we find: RZ² = 2 * RE². Taking the square root of both sides, we get: RZ = √(2 * RE²).

Therefore, the radius of Zeta is approximately √2 times the radius of Earth.

User Anders Johansson
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2 votes

Answer:

1/ square root of 2

Step-by-step explanation:

Inverse proportionality of it is equal to square root of the of gravitational field

User ITChristian
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