Question

(1) A 190-kg object and a 490-kg object are separated by 4.40 m.

(a) Find the magnitude of the net gravitational force exerted by these objects on a 63.0-kg object placed midway between them. [N]

(b) At what position (other than an infinitely remote one) can the 63.0-kg object be placed so as to experience a net force of zero from the other two objects? [m from the 490 kg mass toward the 190 kg mass ]

(2)

(a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape the solar system if it starts at the Earth's orbit? [km/s]

(b) Voyager 1 achieved a maximum speed of 125,000 km/h on its way to photograph Jupiter. Beyond what distance from the Sun is this speed sufficient to escape the solar system? [m]

I really need help with these 2 questions! Just need a final answer, little explanation required but please include the numerical work so I can understand and also please follow the units in the [brackets]

Answer 1

Force = G * M1 * M2 /r^2

When the 63 kg particle feels the same force of attration to the 190 and 490 kg objects, the 63 kg object is in equlibrium, so net force = 0 N


G * 490 * 63 /r1^2 = G * 190 * 63/r2^2

490/r1^2 = 190/r2^2

490 * r2^2 = 190 * r1^2

r2^2/ r1^2 = 190/490 = 19/49

r2/r1 = (19/49)^0.5

r2 = r1 * (19/49)^0.5

The is ratio of distances.

The distance to the 490 object is (19/49)^0.5 * the distance to the 190 kg object

If the 190-kg object and a 490-kg object are separated by 4.40 m, r1 + r2 = 4.40


Now you have 2 equations in 2 variables, solve for r1and r2
r1 + r2 = 4.40
r2 = r1 * (19/49)^0.55

2-

To compute escape velocity we are equating the potential energy Pe an object of mass m can have if achieved an orbit of radius R around a mass M to the kinetic energy Ke it could achieve if allowed to free fall back to body M.

Ke=Pe
0.5 m V^2 = FR
F= GMm/R^2 we have
0.5 m V^2 = (GMm/R^2 )R
0.5 m V^2 = GMm/R
V= sqrt(2GMm/(Rm))
V= sqrt(2GM/R)

G- gravitational constant = 6.673 E-11 m3/ (kg s^2)
M- mass of the Sun= 1.9891 E+30 kg
R- radius of the solar system = 7.5x10^15m (He-he...enough to hit Kuiper Belt 30,000 AU and the Oort Cloud 50,000AU. The Oort Cloud has a larger radius, estimated at about 50,000 AU (or 7.5x10^15 m).
Now we are ready

V= sqrt(2GM/R)
V= sqrt(2x6.673 E-11 x 1.9891 E+30 /7.5x10^15 )
V=188 m/s
V= 680 km/h
If you are looking to establish an orbit around the Sun escape velocity of of 620,000m/s would do just fine. The escape velocity is inversely proportional to the ratio of orbit you want to achieve.

"To leave planet Earth an escape velocity of 11.2 km/s is required, however a speed of 42.1 km/s is required to escape the Sun's gravity (and exit the solar system) from the same position"

b-

Voyager 1 achieved a maximum speed of 125 000 km/h on its way to photograph jupiter. Beyond what distance from the sun is this speed sufficient to escape the solar system?

This question is asking, What is the value of R that makes escape velocity equal 125000 km/h? Recall the formula:

V_escape = sqrt(2GM/R)

In this case, V_escape = 125000km/h, and M = mass of sun Ms; and R = distance from sun. So:

125000km/h = sqrt(2GMs/R)

Finally, just solve for R, and plug in known values for G and Ms