Question

In 2005, the space probe Deep Impact launched a 370 kg projectile into Comet Temple 1. Observing the collision helped scientists learn about the comet’s characteristics. The comet is estimated to have a mass of about 9.0 x 10^13 kg.

a. Assuming the estimated mass of the comet at that time was correct, at what distance from the comet’s centre was the gravitational force between the comet and the projectile 32 N?

b. Deep Impact also released a probe to fly by the comet and record images of the collision. Determine the strength of the comet’s gravitational field at the probe’s distance of 5.0 x 103 km from the comet.

Answer 1

Parta a.

Equation: F = G*m1*m2/d^2

Where
F = 32 N
G = 6.67*10^-11 N.m^2/kg^2
m1 = 9.0*10^13kg
m2 =370 kg
d = distance that separate the center of the two objects.

d^2 = G*m1*m2 / F = 6.67*10^-11 N.m^2/kg^2 * 9.0*10^13 kg *370 kg / 32N = 69,409.69 m^2

d = √69,409.69m^2 = 263.5 m

Part B.

The gravitational field of the comet is g = G*m1/d^2

Notice that it does not depend on the mass of other objects.

Notice also that I will use a distance of 5.0 * 10^3 km, because I think that that is the number that you intended to write in the part b. If that is not the number you can put the right number instead because the solution is written step by step.

g = (6.67*10^-11 N*m^2/kg^2)*(9.0*10^13kg)/(5.0*10^3*10^3m)^2 = 2.4*10^-4 N/kg = 2.4*10^-4 m/s^2