An astronaut is on the surface of a new planet that has a radius of 6.1×10^6m (similar to Venus) and a mass of 6.4×10^23kg (similar to Mars). If she dropped her cell phone, what would be the free-fall acceleration of the phone? Recall that G=6.67×10^−11 N⋅m2/kg2.
Express your answer to two significant figures and indicate the appropriate units.
Well, if the astronaut dropped her cell phone, it would start to experience the gravitational pull of the new planet. To find the free-fall acceleration of the phone, we can use the formula:
g = G * (M / r^2),
where g is the acceleration due to gravity, G is the gravitational constant, M is the mass of the planet, and r is the radius of the planet.
Plugging in the values, we have:
g = (6.67×10^−11 N⋅m^2/kg^2) * (6.4×10^23 kg) / (6.1×10^6 m)^2.
Calculating this out gives us g ≈ 9.7 m/s^2.
So, the free-fall acceleration of the cell phone on the new planet would be approximately 9.7 m/s^2. I hope it has a good protective case to handle that fall!
To find the free-fall acceleration of an object on the surface of a planet, we can use the formula:
a = G * (M / R^2)
where:
a is the free-fall acceleration,
G is the gravitational constant (6.67×10^−11 N⋅m^2/kg^2),
M is the mass of the planet (6.4×10^23 kg), and
R is the radius of the planet (6.1×10^6 m).
Substituting the given values into the formula:
a = (6.67×10^−11 N⋅m^2/kg^2) * (6.4×10^23 kg) / (6.1×10^6 m)^2
Calculating this expression will give us the free-fall acceleration of the phone on the surface of the planet.
To find the free-fall acceleration of the cell phone on the surface of the new planet, we can use the formula for the acceleration due to gravity:
acceleration due to gravity (g) = (Gravitational constant * mass of the planet) / (radius of the planet)^2
Given that the Gravitational constant G is 6.67×10^−11 N⋅m^2/kg^2, the mass of the planet is 6.4×10^23 kg, and the radius of the planet is 6.1×10^6 m, we can substitute these values into the formula:
g = (6.67×10^−11 N⋅m^2/kg^2 * 6.4×10^23 kg) / (6.1×10^6 m)^2
Now, let's calculate the value:
g = (6.67×10^−11 N⋅m^2/kg^2 * 6.4×10^23 kg) / (6.1×10^6 m)^2
≈ 1.102 m/s^2
Therefore, the free-fall acceleration of the cell phone on the surface of the new planet is approximately 1.10 m/s^2.