Two baseballs, each with a mass of 0.157 kg, are separated by a distance of 379 m in outer space, far from any other objects. If the balls are released from rest, what speed do they have when their separation has decreased to 173 m? Ignore the gravitational effects from any other objects.

Question

Two baseballs, each with a mass of 0.157 kg, are separated by a distance of 379 m in outer space, far from any other objects. If the balls are released from rest, what speed do they have when their separation has decreased to 173 m? Ignore the gravitational effects from any other objects.

What would the speed be, if the mass of the balls is quadrupled?

I tried using the equation v^2 = G M ((1 / R2) - (1 / R1)) and got the answer 2.07x10^-7 m/s, but the homework says that the answer is wrong.

Answer 1

Its wrong.

Answer 2

To find the speed of the baseballs as their separation decreases, we can use the conservation of mechanical energy. Since only gravitational potential energy is involved, we can equate the initial potential energy to the final kinetic energy.

The potential energy at the initial separation is given by:
PE_initial = -(G * m^2) / r_initial

The kinetic energy at the final separation is given by:
KE_final = (1/2) * m * v_final^2

Setting these two equal, we have:
-(G * m^2) / r_initial = (1/2) * m * v_final^2

Simplifying and rearranging the equation, we can solve for v_final:
v_final = sqrt((2 * G * m^2) / r_initial)

Using the values given in the problem:
m = 0.157 kg (mass of each baseball)
r_initial = 379 m (initial separation)

Plugging in these values, we find:
v_final = sqrt((2 * G * (0.157 kg)^2) / 379 m)

Now, let's calculate the speed using this equation:

Step 1: Calculate G:
G = 6.67430 * 10^-11 m^3 / (kg * s^2)

Step 2: Plug in the values:
v_final = sqrt((2 * 6.67430 * 10^-11 m^3 / (kg * s^2) * (0.157 kg)^2) / 379 m)

Evaluating the expression, we find:
v_final ≈ 8.77 x 10^-8 m/s

Now, let's calculate the speed when the mass of the balls is quadrupled. If the mass is quadrupled, the new mass would be 4 times the original mass, which is 4 * 0.157 kg = 0.628 kg.

Using the same equation, with the new mass value:
v_final = sqrt((2 * G * (0.628 kg)^2) / 379 m)

Plug in the values and evaluate the expression:
v_final ≈ 1.75 x 10^-7 m/s

So, when the mass of the balls is quadrupled, the speed would be approximately 1.75 x 10^-7 m/s.

Answer 3

PE2-PE1 = 2KE

G•m^2/r^2 - G•m^2/R^2 = 2•m•v^2/2.
v=sqrt[G•m•(R^2 - r^2)]/r•R =
= 1.66•10^-8 m/s

Answer 4

v = sqrt(G.157((1/173) - (1/379))) = 1.81x10^-7 m/s

For the second question just multiply the mass by 4.
v= 3.63x10^-7 m/s

Two baseballs, each with a mass of 0.157 kg, are separated by a distance of 379 m in outer space, far from any other objects. If the balls are released from rest, what speed do they have when their separation has decreased to 173 m? Ignore the gravitational effects from any other objects.

Its wrong.

To find the speed of the baseballs as their separation decreases, we can use the conservation of mechanical energy. Since only gravitational potential energy is involved, we can equate the initial potential energy to the final kinetic energy.

PE2-PE1 = 2KE

v = sqrt(G*.157*((1/173) - (1/379))) = 1.81x10^-7 m/s

v = sqrt(G.157((1/173) - (1/379))) = 1.81x10^-7 m/s