Question

A planet orbiting a distant star has radius 4.14×106 m . The escape speed for an object launched from this planet's surface is 5.15×103 m/s .

What is the acceleration due to gravity at the surface of the planet?

Answer 1

The acceleration due to gravity on the surface of the planet with a radius of 4.14×10¶ m and escape speed of 5.15×10³ m/s is 3.20 m/s².

Step-by-step explanation:

To calculate the acceleration due to gravity at the surface of a planet, you can use the formula for escape velocity ⋅ve⋅, which is ve = √(2⋅G⋅M/r), where G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet. Rearranging the formula to solve for the acceleration due to gravity (g), we get g = ve² / (2⋅r).

Given the escape speed ve = 5.15×10³ m/s and the planet radius r = 4.14×10¶ m, we can plug these values into the formula:

g = (5.15×10³ m/s)² / (2⋅(4.14×10¶ m))

g = 26.5225×10¶ m²/s² / (8.28×10¶ m)

g = 3.20 m/s²

Therefore, the acceleration due to gravity on the surface of the planet is 3.20 m/s².

Answer 2

The acceleration due to gravity at the surface of a planet can be calculated using the escape speed and radius of the planet.

Step-by-step explanation:

The acceleration due to gravity at the surface of a planet can be calculated using the formula g = GM/r^2, where G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet.

In this case, the escape speed of the planet is given as 5.15x10^3 m/s. The escape speed can be related to the acceleration due to gravity using the formula v_escape = sqrt(2gR), where R is the radius of the planet.

Solving for g, we can substitute the values given for the escape speed and radius of the planet to find the acceleration due to gravity at the surface.